力扣--LCR 129.字母迷宫

题目

字母迷宫游戏初始界面记作 m x n 二维字符串数组 grid，请判断玩家是否能在 grid 中找到目标单词 target。

注意：寻找单词时必须按照字母顺序，通过水平或垂直方向相邻的单元格内的字母构成，同时，同一个单元格内的字母不允许被重复使用。

示例 1：

输入：grid = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], target = "ABCCED"

输出：true

示例 2：

输入：grid = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], target = "SEE"

输出：true

示例 3：

输入：grid = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], target = "ABCB"

输出：false

提示：

复制代码

m == grid.length
n = grid[i].length
1 <= m, n <= 6
1 <= target.length <= 15
grid 和 target 仅由大小写英文字母组成

代码

class Solution {

int n;

int m;

int len;

boolean[][] visited;

public boolean wordPuzzle(char[][] grid, String target) {

this.n = grid.length;

this.m = grid[0].length;

this.len = target.length();

visited = new boolean[n][m];

复制代码

    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            if(dsf(grid, i, j, target, 0)){
                return true;
            }
        }
    }

    return false;
}
boolean dsf(char[][] grid, int i, int j, String target, int k){
    // 判断是否越界，已经访问过或者不匹配
    if(i < 0 || i >= n || j < 0 || j >= m || grid[i][j] != target.charAt(k)){
        return false;
    }

    if(k == len - 1){
        return true;
    }
    //visited[i][j] = true;
    grid[i][j] = '\n';
    // 四个方向都搜索一下
    boolean res = dsf(grid, i, j + 1, target, k + 1) ||
                  dsf(grid, i + 1, j, target, k + 1) ||
                  dsf(grid, i, j - 1, target, k + 1) ||
                  dsf(grid, i - 1, j, target, k + 1);

    grid[i][j] = target.charAt(k);
    return res;
    // 时间复杂度：mn * 3^len

}

时间复杂度：O(3^K * MN) ：最差情况下，需要遍历矩阵中长度为 KK 字符串的所有方案，时间复杂度为 O(3^K)；矩阵中共有 MN 个起点，时间复杂度为 O(MN)

空间复杂度：O(k)