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‎README.adoc

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//|{doc_base_url}/0914-x-of-a-kind-in-a-deck-of-cards.adoc[题解]
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//|Easy
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/partition-array-into-disjoint-intervals/[915. Partition Array into Disjoint Intervals^]
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//|{source_base_url}/_0915_PartitionArrayIntoDisjointIntervals.java[Java]
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//|{doc_base_url}/0915-partition-array-into-disjoint-intervals.adoc[题解]
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//|Medium
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/partition-array-into-disjoint-intervals/[915. Partition Array into Disjoint Intervals^]
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|{source_base_url}/_0915_PartitionArrayIntoDisjointIntervals.java[Java]
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|{doc_base_url}/0915-partition-array-into-disjoint-intervals.adoc[题解]
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|Medium
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//|{counter:codes}
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//|{leetcode_base_url}/word-subsets/[916. Word Subsets^]
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//|{source_base_url}/_0916_WordSubsets.java[Java]

‎docs/0915-partition-array-into-disjoint-intervals.adoc

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[#0915-partition-array-into-disjoint-intervals]
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= 915. Partition Array into Disjoint Intervals
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= 915. 分割数组
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{leetcode}/problems/partition-array-into-disjoint-intervals/[LeetCode - Partition Array into Disjoint Intervals^]
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https://leetcode.cn/problems/partition-array-into-disjoint-intervals/[LeetCode - 915. 分割数组 ^]
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Given an array `A`, partition it into two (contiguous) subarrays `left` and `right` so that:
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给定一个数组 `nums` ，将其划分为两个连续子数组 `left` 和 `right`， 使得：
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* `left` 中的每个元素都小于或等于 `right` 中的每个元素。
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* `left` 和 `right` 都是非空的。
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* `left` 的长度要尽可能小。
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* Every element in `left` is less than or equal to every element in `right`.
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* `left` and `right` are non-empty.
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* `left` has the smallest possible size.
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_在完成这样的分组后返回 `left` 的 *长度 *_。
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用例可以保证存在这样的划分方法。
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Return the *length* of `left` after such a partitioning. It is guaranteed that such a partitioning exists.
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*示例 1：*
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....
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输入：nums = [5,0,3,8,6]
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输出：3
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解释：left = [5,0,3]，right = [8,6]
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....
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*Example 1:*
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*示例 2：*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [5,0,3,8,6]
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*Output:* 3
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*Explanation:* left = [5,0,3], right = [8,6]
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----
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*Example 2:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [1,1,1,0,6,12]
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*Output:* 4
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*Explanation:* left = [1,1,1,0], right = [6,12]
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----
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....
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输入：nums = [1,1,1,0,6,12]
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输出：4
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解释：left = [1,1,1,0]，right = [6,12]
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....
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*提示：*
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*Note:*
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* `2 \<= nums.length \<= 10^5^`
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* `0 \<= nums[i] \<= 10^6^`
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* 可以保证至少有一种方法能够按题目所描述的那样对 `nums` 进行划分。
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* `2 <= A.length <= 30000`
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* `0 <= A[i] <= 10^6`
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* It is guaranteed there is at least one way to partition `A` as described.
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== 思路分析
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两次遍历，先从右向左查找最小值，然后从左向右查找最大值，当最大值小于下一个的最小值，则就到了分界线。
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下面这个题解更简单，只需要遍历不到一遍：
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image::images/0915-10.png[{image_attr}]
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[[src-0915]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0915_PartitionArrayIntoDisjointIntervals.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0915_PartitionArrayIntoDisjointIntervals_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/partition-array-into-disjoint-intervals/solutions/1913934/fen-ge-shu-zu-by-leetcode-solution-t4pm/[915. 分割数组 - 官方题解^]
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. https://leetcode.cn/problems/partition-array-into-disjoint-intervals/solutions/1920768/zhua-wa-mou-si-tu-jie-leetcode-by-muse-7-omut/[915. 分割数组 - 【爪哇缪斯】图解LeetCode^] -- 不到一遍的遍历，牛逼！

‎docs/images/0915-10.png

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‎docs/index.adoc

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// include::0915-partition-array-into-disjoint-intervals.adoc[leveloffset=+1]
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include::0915-partition-array-into-disjoint-intervals.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{doc_base_url}/1022-sum-of-root-to-leaf-binary-numbers.adoc[题解]
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|✅ 深度优先遍历。
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|{counter:codes2503}
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|{leetcode_base_url}/partition-array-into-disjoint-intervals/[915. 分割数组^]
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|{doc_base_url}/0915-partition-array-into-disjoint-intervals.adoc[题解]
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|✅ 两遍遍历。有题解可以不到一遍遍历即可解决问题。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0915_PartitionArrayIntoDisjointIntervals.java

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package com.diguage.algo.leetcode;
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public class _0915_PartitionArrayIntoDisjointIntervals {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-05-26 11:39:29
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*/
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public int partitionDisjoint(int[] nums) {
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int length = nums.length;
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int[] ext = new int[length];
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ext[length - 1] = nums[length - 1];
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int min = nums[length - 1];
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for (int i = length - 2; i >= 0; i--) {
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min = Math.min(min, nums[i]);
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ext[i] = min;
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}
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int max = nums[0];
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for (int i = 0; i < length - 1; i++) {
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max = Math.max(max, nums[i]);
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if (max <= ext[i + 1]) {
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return i + 1;
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}
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}
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return length - 1;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0915_PartitionArrayIntoDisjointIntervals()
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.partitionDisjoint(new int[]{1, 1, 1, 0, 6, 12});
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}
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}

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