Question:

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.

题目大意是在矩阵中找是否有word的路径，上下左右都可以

Algorithm:

深度优先搜索，上下左右，用一个矩阵记录该列该行的元素是否被搜索过

Accepted Code：

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int M=board.size();
        int N=board[0].size();
        vector<vector<int>> v(M,vector<int> (N));
        for(int i=0;i<M;i++)
            for(int j=0;j<N;j++)
            {
                if(board[i][j]==word[0])
                    if(dfs(board,word,i,j,0,v))
                        return true;
            }
        return false;
    }
    bool dfs(vector<vector<char>>& board, string word,int x,int y,int index,vector<vector<int>>& v)
    {
        int M=board.size();
        int N=board[0].size();
        if(index==word.size()-1)
            return true;
        v[x][y]=1;
        if(x+1<M&&v[x+1][y]==0&&board[x+1][y]==word[index+1])   //right
            if(dfs(board,word,x+1,y,index+1,v))
                return true;
        if(x-1>=0&&v[x-1][y]==0&&board[x-1][y]==word[index+1])  //left
            if(dfs(board,word,x-1,y,index+1,v))
                return true;
        if(y+1<N&&v[x][y+1]==0&&board[x][y+1]==word[index+1])   //up
            if(dfs(board,word,x,y+1,index+1,v))
                return true;
        if(y-1>=0&&v[x][y-1]==0&&board[x][y-1]==word[index+1])  //down
            if(dfs(board,word,x,y-1,index+1,v))
                return true;
        v[x][y]=0;       //如果上下左右都走不下去了，那么之前的都归0
        return false;
    }
};