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用二进制表示获得的钥匙，假设n=钥匙个数

000000000代表没有钥匙，0000000001代表有idx为1的钥匙，0000000011代表有idx=1，2的钥匙

（这方法巧妙又复杂..

代码：

class Solution {static int[][] dirs = {{-1,0},{1,0},{0,-1},{0,1}};public int shortestPathAllKeys(String[] grid) {int m = grid.length, n = grid[0].length();int startx = 0,starty = 0;Map<Character, Integer> keyToIndex = new HashMap<>();//存钥匙的字母和对应的idx序号for(int i=0;i<grid.length;i++){for(int j=0;j<grid[i].length();j++){if(grid[i].charAt(j)=='@'){startx = i;starty = j;}else if(Character.isLowerCase(grid[i].charAt(j))){if(!keyToIndex.containsKey(grid[i].charAt(j))){int idx = keyToIndex.size();keyToIndex.put(grid[i].charAt(j), idx);}                    }}}Queue<int[]> queue = new ArrayDeque<int[]>();int[][][] dist = new int[m][n][1<<keyToIndex.size()];//第三维是2的size次方 列举钥匙的所有可能for(int i=0;i<m;i++){for(int j=0;j<n;j++){Arrays.fill(dist[i][j],-1);}} queue.offer(new int[]{startx,starty,0});dist[startx][starty][0] = 0;while(!queue.isEmpty()){int[] arr = queue.poll();int x = arr[0],y = arr[1],mask = arr[2];//mask是钥匙的排列for(int i=0;i<4;i++){int nx = x + dirs[i][0];int ny = y + dirs[i][1];if(nx>=0 && nx<m && ny>=0 && ny<n &&grid[nx].charAt(ny)!='#'){if(grid[nx].charAt(ny)=='.'||grid[nx].charAt(ny)=='@'){if(dist[nx][ny][mask] == -1){dist[nx][ny][mask] = dist[x][y][mask]+1;queue.offer(new int[]{nx,ny,mask});}}else if(Character.isLowerCase(grid[nx].charAt(ny))){int idx = keyToIndex.get(grid[nx].charAt(ny));if(dist[nx][ny][mask|(1<<idx)] == -1){dist[nx][ny][mask|(1<<idx)] = dist[x][y][mask]+1;if((mask|(1<<idx))==(1<<keyToIndex.size())-1){return dist[nx][ny][mask|(1<<idx)];}queue.offer(new int[]{nx,ny,mask|(1<<idx)});}}else{int idx = keyToIndex.get(Character.toLowerCase(grid[nx].charAt(ny)));if((mask&(1<<idx))!=0&&dist[nx][ny][mask]==-1){dist[nx][ny][mask] = dist[x][y][mask]+1;queue.offer(new int[]{nx,ny,mask});}}}}}  return -1;     }
}