Leetcode_c++： Word Search (079）

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.

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算法

dfs
O（n^2m^2）

题意很简单，给你一个二维字母的数组，可以上下左右走，查找是否某个单词是否存在。同一位置的字母不可以被使用多次。

解题思路： 类似于迷宫，递归回溯。需要一个辅助数组记录走过的位置，防止同一个位置被使用多次。

http://www.cnblogs.com/ganganloveu/p/4188131.html

---

class Solution {private:    bool dfs(vector<vector<char> > &board,int x,int y,string word,int idx,             vector<vector<bool> > &vis){        if(idx==word.length())            return true;        if(x<0 ||x>=board.size()||y<0||y>=board[0].size()||vis[x][y]||           board[x][y]!=word[idx])            return false;        vis[x][y]=true;        bool ret=        dfs(board,x+1,y,word,idx+1,vis)||        dfs(board,x-1,y,word,idx+1,vis)||        dfs(board,x,y+1,word,idx+1,vis)||        dfs(board,x,y-1,word,idx+1,vis);        vis[x][y]=false;        return ret;    }public:    bool exist(vector<vector<char> >& board, string word) {        if(board.empty()||board[0].empty())            return word=="";        int n=board.size(),m=board[0].size();        vector<vector<bool> >  vis(n,vector<bool>(m,false));        for(int i=0;i<n;++i)            for(int j=0;j<m;++j)                if(dfs(board,i,j,word,0,vis))                    return true;        return false;    }};