34. Word Search FROM Leetcode

题目

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 =

[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

频度： 4

解题之法

class Solution {
private:
    bool dfs(vector<vector<char>>& board, int row, int col, const string &word, int start, int M, int N, int sLen)
    {
        char curC;
        bool res = false;
        if( (curC = board[row][col]) != word[start]) return false;
        if(start==sLen-1) return true;
        board[row][col] = '*';
        if(row>0) res = dfs(board, row-1, col, word, start+1, M, N, sLen);
        if(!res && row < M-1) res = dfs(board, row+1, col, word, start+1, M, N, sLen);
        if(!res && col > 0)   res = dfs(board, row, col-1, word, start+1, M, N, sLen);
        if(!res && col < N-1) res = dfs(board,  row, col+1, word, start+1, M, N, sLen);
        board[row][col] = curC;
        return res;
    }
    
public:
    bool exist(vector<vector<char>>& board, string word) {
        int M,N,i,j,sLen = word.size();
        if( (M=board.size()) && (N=board[0].size()) && sLen)
        {
            for(i=0; i<M; ++i)
                for(j=0; j<N; ++j)
                    if(dfs(board, i, j, word, 0, M, N, sLen)) return true;
        }
        return false;
    }
};