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增加解题范式，增加注释
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‎docs/0000-00-note.adoc

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== TODO
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* [ ] https://leetcode.cn/problems/number-of-islands/solutions/211211/dao-yu-lei-wen-ti-de-tong-yong-jie-fa-dfs-bian-li-/[200. 岛屿数量 - 岛屿类问题的通用解法、DFS 遍历框架^]
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== 解题范式
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. Two Pointers
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.. Pair with Target Sum (easy)
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.. Find Non-Duplicate Number Instances (easy)
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.. Squaring a Sorted Array (easy)
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.. Triplet Sum to Zero (medium)
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.. Triplet Sum Close to Target (medium)
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.. Triplets with Smaller Sum (medium)
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.. Subarrays with Product Less than a Target (medium)
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.. Dutch National Flag Problem (medium)
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.. Problem Challenge 1: Quadruple Sum to Target (medium)
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.. Problem Challenge 2: Comparing Strings containing Backspaces (medium)
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.. Problem Challenge 3: Minimum Window Sort (medium)
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. Fast & Slow Pointers
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.. LinkedList Cycle (easy)
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.. Middle of the LinkedList (easy)
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.. Start of LinkedList Cycle (medium)
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.. Happy Number (medium)
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.. Problem Challenge 1: Palindrome LinkedList (medium)
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.. Problem Challenge 2: Rearrange a LinkedList (medium)
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.. Problem Challenge 3: Cycle in a Circular Array (hard)
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. Sliding Window
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.. Maximum Sum Subarray of Size K (easy)
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.. Smallest Subarray With a Greater Sum (easy)
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.. Longest Substring with K Distinct Characters (medium)
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.. Fruits into Baskets (medium)
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.. Longest Substring with Same Letters after Replacement (hard)
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.. Longest Subarray with Ones after Replacement (hard)
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.. Problem Challenge 1: Permutation in a String (hard)
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.. Problem Challenge 2: String Anagrams (hard)
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.. Problem Challenge 3: Smallest Window containing Substring (hard)
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.. Problem Challenge 4: Words Concatenation (hard)
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. Merge Intervals
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.. Merge Intervals (medium)
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.. Insert Interval (medium)
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.. Intervals Intersection (medium)
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.. Conflicting Appointments (medium)
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.. Problem Challenge 1: Minimum Meeting Rooms (hard)
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.. Problem Challenge 2: Maximum CPU Load (hard)
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.. Problem Challenge 3: Employee Free Time (hard)
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. Cyclic Sort
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.. Cyclic Sort (easy)
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.. Find the Missing Number (easy)
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.. Find all Missing Numbers (easy)
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.. Find the Duplicate Number (easy)
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.. Find all Duplicate Numbers (easy)
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.. Problem Challenge 1: Find the Corrupt Pair (easy)
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.. Problem Challenge 2: Find the Smallest Missing Positive Number (medium)
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.. Problem Challenge 3: Find the First K Missing Positive Numbers (hard)
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. In-place Reversal of a Linked List
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.. Reverse a LinkedList (easy)
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.. Reverse a Sub-list (medium)
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.. Reverse every K-element Sub-list (medium)
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.. Problem Challenge 1: Reverse alternating K-element Sub-list (medium)
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.. Problem Challenge 2: Rotate a LinkedList (medium)
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. Stacks
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.. Operations on Stack
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.. Implementing Stack Data Structure
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.. Applications of Stack
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.. Problem 1: Balanced Parentheses
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.. Problem 2: Reverse a String
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.. Problem 3: Decimal to Binary Conversion
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.. Problem 4: Next Greater Element
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.. Problem 5: Sorting a Stack
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.. Problem 6: Simplify Path
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. Monotonic Stack
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.. Remove Nodes From Linked List (medium)
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.. Remove All Adjacent Duplicates In String (easy)
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.. Next Greater Element (easy)
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.. Daily Temperatures (easy)
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.. Remove All Adjacent Duplicates in String II (medium)
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.. Remove K Digits (hard)
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. Hash Maps
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.. Introduction to Hash Tables
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.. Issues with Hash Tables
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.. Problem 1: First Non-repeating Character (easy)
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.. Problem 2: Largest Unique Number (easy)
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.. Problem 3: Maximum Number of Balloons (easy)
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.. Problem 4: Longest Palindrome(easy)
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.. Problem 5: Ransom Note (easy)
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. Tree Breadth First Search
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.. Binary Tree Level Order Traversal (easy)
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.. Reverse Level Order Traversal (easy)
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.. Zigzag Traversal (medium)
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.. Level Averages in a Binary Tree (easy)
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.. Minimum Depth of a Binary Tree (easy)
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.. Level Order Successor (easy)
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.. Connect Level Order Siblings (medium)
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.. Problem Challenge 1: Connect All Level Order Siblings (medium)
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.. Problem Challenge 2: Right View of a Binary Tree (easy)
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. Tree Depth First Search
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.. Binary Tree Path Sum (easy)
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.. All Paths for a Sum (medium)
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.. Sum of Path Numbers (medium)
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.. Path With Given Sequence (medium)
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.. Count Paths for a Sum (medium)
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.. Problem Challenge 1: Tree Diameter (medium)
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.. Problem Challenge 2: Path with Maximum Sum (hard)
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. Graphs
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.. Graph Representations
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.. Graph as an Abstract Data Type (ADT)
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.. Graph Traversal - Depth First Search(DFS)
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.. Graph Traversal - Breadth First Search (BFS)
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.. Problem 1: Find if Path Exists in Graph(easy)
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.. Problem 2: Number of Provinces (medium)
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.. Problem 3: Minimum Number of Vertices to Reach All Nodes(medium)
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. Island (Matrix Traversal)
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.. Number of Islands (easy)
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.. Biggest Island (easy)
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.. Flood Fill (easy)
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.. Number of Closed Islands (easy)
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.. Problem Challenge 1 (easy)
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.. Problem Challenge 2 (medium)
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.. Problem Challenge 3 (medium)
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. Two Heaps
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.. Find the Median of a Number Stream (medium)
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.. Sliding Window Median (hard)
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.. Maximize Capital (hard)
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.. Problem Challenge 1: Next Interval (hard)
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. Subsets
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.. Subsets (easy)
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.. Subsets With Duplicates (easy)
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.. Permutations (medium)
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.. String Permutations by changing case (medium)
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.. Balanced Parentheses (hard)
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.. Unique Generalized Abbreviations (hard)
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.. Problem Challenge 1: Evaluate Expression (hard)
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.. Problem Challenge 2: Structurally Unique Binary Search Trees (hard)
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.. Problem Challenge 3: Count of Structurally Unique Binary Search Trees (hard)
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. Modified Binary Search
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.. Order-agnostic Binary Search (easy)
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.. Ceiling of a Number (medium)
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.. Next Letter (medium)
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.. Number Range (medium)
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.. Search in a Sorted Infinite Array (medium)
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.. Minimum Difference Element (medium)
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.. Bitonic Array Maximum (easy)
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.. Problem Challenge 1: Search Bitonic Array (medium)
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.. Problem Challenge 2: Search in Rotated Array (medium)
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.. Problem Challenge 3: Rotation Count (medium)
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. Bitwise XOR
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.. Single Number (easy)
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.. Two Single Numbers (medium)
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.. Complement of Base 10 Number (medium)
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.. Problem Challenge 1: Flip and Invert an Image (hard)
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. Top 'K' Elements
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.. Top 'K' Numbers (easy)
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.. Kth Smallest Number (easy)
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.. 'K' Closest Points to the Origin (easy)
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.. Connect Ropes (easy)
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.. Top 'K' Frequent Numbers (medium)
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.. Frequency Sort (medium)
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.. Kth Largest Number in a Stream (medium)
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.. 'K' Closest Numbers (medium)
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.. Maximum Distinct Elements (medium)
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.. Sum of Elements (medium)
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.. Rearrange String (hard)
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.. Problem Challenge 1: Rearrange String K Distance Apart (hard)
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.. Problem Challenge 2: Scheduling Tasks (hard)
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.. Problem Challenge 3: Frequency Stack (hard)
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. Greedy Algorithms
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.. Valid Palindrome II (easy)
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.. Maximum Length of Pair Chain (medium)
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.. Minimum Add to Make Parentheses Valid (medium)
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.. Remove Duplicate Letters (medium)
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.. Largest Palindromic Number (medium)
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.. Removing Minimum and Maximum From Array (medium)
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. 0/1 Knapsack (Dynamic Programming)
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.. 0/1 Knapsack (medium)
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.. Equal Subset Sum Partition (medium)
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.. Subset Sum (medium)
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.. Minimum Subset Sum Difference (hard)
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.. Problem Challenge 1: Count of Subset Sum (hard)
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.. Problem Challenge 2: Target Sum (hard)
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. Backtracking
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.. Combination Sum (medium)
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.. Word Search (medium)
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.. Factor Combinations (medium)
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.. Split a String Into the Max Number of Unique Substrings (medium)
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.. Sudoku Solver (hard)
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. Trie
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.. Implement Trie (Prefix Tree) (medium)
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.. Index Pairs of a String (easy)
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.. Design Add and Search Words Data Structure (medium)
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.. Extra Characters in a String (medium)
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.. Search Suggestions System (medium)
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. Topological Sort (Graph)
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.. Topological Sort (medium)
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.. Tasks Scheduling (medium)
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.. Tasks Scheduling Order (medium)
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.. All Tasks Scheduling Orders (hard)
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.. Alien Dictionary (hard)
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.. Problem Challenge 1: Reconstructing a Sequence (hard)
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.. Problem Challenge 2: Minimum Height Trees (hard)
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. Union Find
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.. Redundant Connection (medium)
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.. Number of Provinces (medium)
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.. Is Graph Bipartite? (medium)
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.. Path With Minimum Effort (medium)
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. Ordered Set
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.. Merge Similar Items (easy)
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.. 132 Pattern (medium)
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.. My Calendar I (medium)
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.. Longest Continuous Subarray (medium)
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. Prefix Sum
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.. Find the Middle Index in Array (easy)
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.. Left and Right Sum Differences (easy)
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.. Maximum Size Subarray Sum Equals k (medium)
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.. Binary Subarrays With Sum (medium)
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.. Subarray Sums Divisible by K (medium)
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.. Sum of Absolute Differences in a Sorted Array (medium)
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.. Subarray Sum Equals K (medium)
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. Multi-threaded
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.. Same Tree (medium)
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.. Invert Binary Tree (medium)
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.. Binary Search Tree Iterator (medium)
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== 参考资料
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. https://www.designgurus.io/course/grokking-the-coding-interview[Grokking the Coding Interview Patterns^]

‎docs/0000-23-monotonic-stack.adoc

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一般会用到 `Deque` 的以下四个方法：
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* `deque.isEmpty()`：如果deque不包含任何元素，则返回true，否则返回false。因为要栈顶元素在满足要求的时候要弹出，所以需要进行空栈判断。有些场景，可能栈一定不会空的时候，就不需要该方法进行空栈判断。
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* `deque.push(e)`：将元素e入栈。一定能用到该方法。
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* `deque.pop()`：将栈顶元素弹出，并返回当前弹出的栈顶元素。一定能用到该方法。
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* `deque.peek()`：获取栈顶元素，不弹出。一定能用到该方法。
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* `deque.isEmpty()`：如果 `deque` 不包含任何元素，则返回 `true`，否则返回 `false`。因为要栈顶元素在满足要求的时候要弹出，所以需要进行空栈判断。有些场景，可能栈一定不会空的时候，就不需要该方法进行空栈判断。
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* `deque.push(e)`：将元素 `e` 入栈。
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* `deque.pop()`：将栈顶元素弹出，并返回当前弹出的栈顶元素。
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* `deque.peek()`：获取栈顶元素，不弹出。
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[{java_src_attr}]
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----
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// 第一个一个单调栈
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// 定义一个单调栈
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Deque<Integer> stack = new LinkedList<>();
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// 第一个元素，直接添加
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stack.push(0); // 这里存的是数组下标
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for (int i = 1; i < nums.length; i++) {
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// 单调递减栈这里就是大于，即 nums[i] > nums[deque.peek()]
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// 单调递增栈这里就是大于，即 nums[i] > nums[deque.peek()]
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if (nums[i] < nums[stack.peek()]) {
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stack.push(i);
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} else if (nums[i] == nums[stack.peek()]) {
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stack.push(i);
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//此处除了入栈，在有些场景下，还有可能有其他操作
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// 此处除了入栈，在有些场景下，还有可能有其他操作
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// ..............
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} else {
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//循环比较，直到遇到当前元素小于栈顶的元素情况，跳出循环
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// 单调递减栈，这里是小于，即nums[i]<nums[deque.peek()]
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// 循环比较，直到遇到当前元素小于栈顶的元素情况，跳出循环
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// 单调递增栈，这里是小于，即nums[i] < nums[deque.peek()]
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while (!stack.isEmpty() && nums[i] > nums[stack.peek()]) {
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//主要逻辑
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// ............
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// ............
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// 弹出栈顶元素
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stack.pop();
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} stack.push(i);
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}
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stack.push(i);
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}
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}
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----

‎docs/0503-next-greater-element-ii.adoc

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image::images/0503-02.png[{image_attr}]
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// TODO 图解是从前向后处理，代码时从后向前处理，尝试从前向后的处理方案。
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[[src-0503]]
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[{java_src_attr}]
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----

‎docs/index.adoc

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include::1644-lowest-common-ancestor-of-a-binary-tree-ii.adoc[leveloffset=+1]
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include::1905-count-sub-islands.adoc[leveloffset=+1]
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include::0000-00-note.adoc[leveloffset=+1]

‎src/main/java/com/diguage/algo/leetcode/_0503_NextGreaterElementII.java

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}
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int[] result = new int[nums.length];
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Deque<Integer> stack = new LinkedList<>();
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// 只需要将数组“拼接”，遍历两遍数组，就可以解决所有元素后继更大元素的问题
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// 从后向前遍历，再加上单调递增栈，就是时间复杂度为 O(n) 的解决方案
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for (int i = 2 * nums.length - 1; i >= 0; i--) {
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// 取余即可获取当前需要处理的元素
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int index = i % nums.length;
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// 在单调栈不为空的情况下，将栈中小于等于当前元素的值都弹出
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while (!stack.isEmpty() && stack.peek() <= nums[index]) {
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stack.pop();
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}
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// 剩下元素既是比当前元素大的后继元素。为空则是没有更大元素
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// 这里还有一个隐含变量：
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// 由于栈是从后向前添加，则栈顶元素距离当前元素更近。
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// 如果栈不为空，则栈顶元素就是符合条件的元素。
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result[index] = stack.isEmpty() ? -1 : stack.peek();
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stack.push(nums[index]);
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}

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