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12.矩阵中的路径 (Medium)*

题目描述*

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*

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

输出：true

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

输出：false

提示*

1 <= board.length <= 200, 1 <= board[i].length <= 200

代码*

DFS，回溯，脑子不好使，练练练

首先要找到

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(board.size() == 0 || board[0].size() == 0) {
            return false;
        }
        for(int i = 0; i < board.size(); i++) {
            for(int j = 0; j < board[0].size(); j++) {
                if(dfs(board, word, i, j, 0)) {
                    return true;
                }
            }
        }
        return false;
    }
    bool dfs(vector<vector<char>>& board, string& word, int i, int j, int len) {
        if(i >= board.size() || j >= board[0].size() || i < 0 || j < 0 
            || len >= word.length() || word[len] != board[i][j]) {
            return false;
        }
        if(len == word.length() - 1) {
            return true;
        }
        board[i][j] ^= 128;
        if(dfs(board, word, i, j + 1, len + 1)
            || dfs(board, word, i, j - 1, len + 1)
            || dfs(board, word, i + 1, j, len + 1)
            || dfs(board, word, i - 1, j, len + 1)) {
            return true;
        }
        board[i][j] ^= 128;
        return false;
    }

};
最后更新: July 23, 2022