[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.

LeetCode Source

思路：典型的DFS算法，每次从string开始搜寻。

特别注意边界条件，每次搜寻时作标记，如果搜寻没有成功的话，标志消除。

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        for(int i=0;i<board.size();++i){            for(int j=0;j<board[0].size();++j){                if(dfs(board,i,j,word.begin(),word.end()))                    return true;            }        }        return false;    }    bool dfs(vector<vector<char> >&board,int m,int n,string::iterator it,string::iterator end){        if(m<0||n<0||m>board.size()-1||n>board[0].size()-1||board[m][n]!=*it)            return false;        if(board[m][n]==*it&&it+1==end)            return true;        char temp = board[m][n];        board[m][n] = '0'; //做标记    if(dfs(board,m-1,n,it+1,end)||dfs(board,m+1,n,it+1,end)||dfs(board,m,n+1,it+1,end)||dfs(board,m,n-1,it+1,end))        {            return true;        }        board[m][n] = temp; //没有搜寻到标记清楚，没有这条语句会报错        return false;    }};