leetcode 之Work Search

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[  ["ABCE"],  ["SFCS"],  ["ADEE"]]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

深度优先搜索。当board[i][j]==word[0]的时候调用DFS函数。

以下代码来自于 https://github.com/soulmachine/leetcode 

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        const int m = board.size();        const int n = board[0].size();        vector<vector<bool> > visited(m, vector<bool>(n, false));        for (int i = 0; i < m; ++i)            for (int j = 0; j < n; ++j)                if ((board[i][j]==word[0])&&dfs(board, word, 0, i, j, visited))                    return true;        return false;    }    private:    static bool dfs(const vector<vector<char> > &board, const string &word,        int index, int x, int y, vector<vector<bool> > &visited) {        if (index == word.size())             //收敛条件            return true;        if (x < 0 || y < 0 || x >= board.size() || y >= board[0].size())            return false;   //越界，终止条件        if (visited[x][y])           //已经访问过，剪枝            return false;        if (board[x][y] != word[index])       //不相等，剪枝            return false;        visited[x][y] = true;        bool ret = dfs(board, word, index + 1, x - 1, y, visited) || // 上                    dfs(board, word, index + 1, x + 1, y, visited) || // 下                    dfs(board, word, index + 1, x, y - 1, visited) || // 左                    dfs(board, word, index + 1, x, y + 1, visited); // 右        visited[x][y] = false;                                     //回溯        return ret;    }};