https://leetcode.cn/problems/shortest-path-to-get-all-keys

Author: masx200

README.md

@@ -45,6 +45,8 @@ Step 2. Add the dependency
 
 <summary>展开查看</summary>
 
+https://leetcode.cn/problems/shortest-path-to-get-all-keys
+
 https://leetcode.cn/problems/removing-minimum-and-maximum-from-array
 
 https://leetcode.cn/problems/largest-plus-sign

shortest-path-to-get-all-keys/index.ts

@@ -0,0 +1,56 @@
+export default function shortestPathAllKeys(g: string[]): number {
+    const dirs = [
+        [1, 0],
+        [-1, 0],
+        [0, 1],
+        [0, -1],
+    ];
+    const n = g.length,
+        m = g[0].length;
+    let cnt = 0;
+    const dist = new Array<Array<Array<number>>>();
+    for (let i = 0; i < n; i++) {
+        dist[i] = new Array<Array<number>>(m);
+        for (let j = 0; j < m; j++) {
+            dist[i][j] = new Array<number>(1 << 10).fill(Infinity);
+        }
+    }
+    const d: number[][] = [];
+    for (let i = 0; i < n; i++) {
+        for (let j = 0; j < m; j++) {
+            if (g[i][j] == "@") {
+                d.push([i, j, 0]);
+                dist[i][j][0] = 0;
+            } else if (g[i][j] >= "a" && g[i][j] <= "z") cnt++;
+        }
+    }
+    while (d.length > 0) {
+        const info = d.shift() as [number, number, number];
+        const x = info[0],
+            y = info[1],
+            cur = info[2],
+            step = dist[x][y][cur];
+        for (const di of dirs) {
+            const nx = x + di[0],
+                ny = y + di[1];
+            if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
+            const c = g[nx][ny];
+            if (c == "#") continue;
+            if (
+                "A" <= c &&
+                c <= "Z" &&
+                ((cur >> (c.charCodeAt(0) - "A".charCodeAt(0))) & 1) == 0
+            )
+                continue;
+            let ncur = cur;
+            if ("a" <= c && c <= "z") {
+                ncur |= 1 << (c.charCodeAt(0) - "a".charCodeAt(0));
+            }
+            if (ncur == (1 << cnt) - 1) return step + 1;
+            if (step + 1 >= dist[nx][ny][ncur]) continue;
+            d.push([nx, ny, ncur]);
+            dist[nx][ny][ncur] = step + 1;
+        }
+    }
+    return -1;
+}