【剑指Offer】面试题12. 矩阵中的路径

题目

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

思路

同 【LeetCode】79. 单词搜索
以每个位置开头检查是否存在路径。

代码

时间复杂度：O(n * m)
空间复杂度：O(1)

        class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int row = board.size(), col = board[0].size();        
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (board[i][j] == word[0] && dfs(board, i, j, 0, word)) {
                    return true;
                }                
            }
        }
        return false;
    }

    bool dfs(vector<vector<char>> &board, int i, int j, int len, string word) {     
        int row = board.size(), col = board[0].size();
        if (i < 0 || i >= row || j < 0 || j >= col || word[len] != board[i][j]) return false;      
        if (len == word.size() - 1) return true;
        ++len;
        char ch = board[i][j];
        board[i][j] = '#';
        bool ret = dfs(board, i - 1, j, len, word) ||
                   dfs(board, i + 1, j, len, word) ||
                   dfs(board, i, j - 1, len, word) ||
                   dfs(board, i, j + 1, len, word);        
        board[i][j] = ch; //回溯
        return ret;
    }
};
    

另一种写法

使用访问数组表示每个位置是否访问。
时间复杂度：O(n * m)
空间复杂度：O(n * m)

        class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int row = board.size(), col = board[0].size();   
        vector<vector<bool>> visited(row, vector<bool>(col));
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (board[i][j] == word[0] && dfs(board, i, j, 0, word, visited)) {
                    return true;
                }                
            }
        }
        return false;
    }

    bool dfs(vector<vector<char>> &board, int i, int j, int len, string word, vector<vector<bool>> &visited) {     
        int row = board.size(), col = board[0].size();
        if (i < 0 || i >= row || j < 0 || j >= col || visited[i][j] || word[len] != board[i][j]) return false;      
        if (len == word.size() - 1) return true;
        ++len;
        visited[i][j] = true;
        bool ret = dfs(board, i - 1, j, len, word, visited) ||
                   dfs(board, i + 1, j, len, word, visited) ||
                   dfs(board, i, j - 1, len, word, visited) ||
                   dfs(board, i, j + 1, len, word, visited);        
        visited[i][j] = false; //回溯
        return ret;
    }
};
    

原文：https://www.cnblogs.com/galaxy-hao/p/12329118.html