add L0864 solution

Author: ChlZhYa

LeetCode/src/L0864/Solution.java

@@ -0,0 +1,64 @@
+package L0864;
+
+import java.util.*;
+
+public class Solution {
+
+    public int shortestPathAllKeys(String[] grid) {
+        Deque<int[]> queue = new ArrayDeque<>();
+        int keyCnt = 0;
+        int row = grid.length, col = grid[0].length();
+        int[][][] dist = new int[row][col][1 << 10];
+        int[][] dir = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
+        for (int i = 0; i < row; i++) {
+            for (int j = 0; j < col; j++) {
+                Arrays.fill(dist[i][j], 0x3f3f3f3f);
+                char c = grid[i].charAt(j);
+                if (c == '@') {  
+                    queue.add(new int[]{i, j, 0});
+                    dist[i][j][0] = 0;
+                } else if (Character.isLowerCase(c)) {
+                    keyCnt++;
+                }
+            }
+        }
+        while (!queue.isEmpty()) {
+            int[] pos = queue.pollFirst();
+            int x = pos[0], y = pos[1], keyMask = pos[2],step = dist[x][y][keyMask];
+            for (int[] di : dir) {
+                int nx = x + di[0];
+                int ny = y + di[1];
+                if (nx < 0 || nx >= row || ny < 0 || ny >= col) {
+                    continue;
+                }
+                char c = grid[nx].charAt(ny);
+                // 如果是墙，则跳过
+                if (c == '#') {
+                    continue;
+                }
+                // 如果是锁，但是没有对应的钥匙，也跳过
+                if (Character.isUpperCase(c) && ((keyMask >> (c - 'A') & 1) == 0)) {
+                    continue;
+                }
+                int curKeyMask = keyMask;
+                // 如果是钥匙，则扩充当前位置的钥匙数量
+                if (Character.isLowerCase(c)) {
+                    curKeyMask = curKeyMask | 1 << (c - 'a');
+                }
+                // 如果获取了所有钥匙，则返回步数
+                if ((1 << keyCnt) - 1 == curKeyMask) {
+                    return step + 1;
+                }
+                // 如果是其他情况（空格），则继续行走，对应步数 + 1
+                if (step + 1 < dist[nx][ny][curKeyMask]) {
+                    dist[nx][ny][curKeyMask] = step + 1;
+                    queue.addLast(new int[]{nx, ny, curKeyMask});
+                }
+            }
+        }
+        return -1;
+    }
+
+
+
+}