面试题12

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。



示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["a","b"],["c","d"]], word = "abcd"
输出：false

提示：

• 1 <= board.length <= 200

• 1 <= board[i].length <= 200

注意：本题与主站 79 题相同：https://leetcode-cn.com/problems/word-search/

Solutions

1. dfs

class Solution {
public:
    int m, n;
    bool dfs(vector<vector<char>> & board, int i, int j, string & word, int cur) {
        if (cur >= word.size())
            return true;
        if (i < 0 || j < 0 
            || i >= m || j >= n 
            || board[i][j] == '#' 
            || board[i][j] != word[cur])
            return false;
        board[i][j] = '#';
        bool res =  dfs(board, i - 1, j, word, cur + 1)
                 || dfs(board, i + 1, j, word, cur + 1)
                 || dfs(board, i, j - 1, word, cur + 1)
                 || dfs(board, i, j + 1, word, cur + 1);
        board[i][j] = word[cur];
        return res;
    }

    bool exist(vector<vector<char>>& board, string word) {
        if (!word.size()) return false;
        m = board.size(), n = board[0].size();

        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (dfs(board, i, j, word, 0))
                    return true;

        return false;
    }
};