[Leetcode] Word 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.

将已经search过的几位用"."标记，最后没有恢复那些标记，太懒了。。。

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        for(int i=0;i<board.size();i++)        {            for(int j=0;j<board[0].size();j++)            {                if(dfs(board,word,i,j,0,word.length()))                {                    return true;                }            }        }        return false;    }        bool dfs(vector<vector<char> > &board, string word, int i, int j, int depth, int maxDepth)    {        if(depth == maxDepth)        {            return true;        }                if(i<0||j<0) return false;        if(i>=board.size()||j>=board[0].size()) return false;                        if(board[i][j]!=word[depth])        {            return false;        }        else        {            board[i][j] = '.';            if(dfs(board,word,i+1,j,depth+1,maxDepth)) return true;            if(dfs(board,word,i-1,j,depth+1,maxDepth)) return true;            if(dfs(board,word,i,j+1,depth+1,maxDepth)) return true;            if(dfs(board,word,i,j-1,depth+1,maxDepth)) return true;            board[i][j] = word[depth];            return false;        }    }};