[LeetCode]—Word Search

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.

       这道题本身和[LeetCode]-Surrounded Regions 找被包围的点 特别类似。那一题最开始只是遍历外围再递归深入内部四向扩展，而本题是对所有的点进行遍历，然后再根据此点四向扩展。并且本题要求返回true or fasle。并且不能重复访问，则由去重的问题，需要单独用一个数组记录。

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(DFS(board,0,i,j,word,visited))                    return true;            }        return false;    }private:    bool DFS(vector<vector<char> > &board,int index,int x,int y,            string word,vector<vector<bool> > &visited){            if(index==word.size())                return true;                        if(x<0 || y<0 || x>=board.size() || y>=board[0].size())                return false; //越界终止            if(visited[x][y]==true)return false; //访问过，终止            if(board[x][y]!=word[index])return false;            visited[x][y]=true;            bool ret=DFS(board,index+1,x+1,y,word,visited)||                DFS(board,index+1,x-1,y,word,visited)||                DFS(board,index+1,x,y+1,word,visited)||                DFS(board,index+1,x,y-1,word,visited);            visited[x][y]=false;            return ret;    }};