LeetCode 79. 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 =

[  ['A','B','C','E'],  ['S','F','C','S'],  ['A','D','E','E']]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

分析：

回溯＋深度搜索，用数组存放四个方向，依次进行深搜

class Solution {public:    bool exist(vector<vector<char>>& board, string word) {        int n=board.size(),m;        if(n>0)  m=board[0].size();                for(int i=0;i<n;i++){            for(int j=0;j<m;j++){                if(board[i][j]==word[0]){                    memset(vis,0,sizeof(vis));                    vis[i][j]=1;                    dfs(1,i,j,board,word,n,m);                    if(flag) return flag;                }            }        }        return false;    }private:    int dic[4][2]={1,0,-1,0,0,1,0,-1};    int vis[1005][1005];    bool flag=false;    void dfs(int index,int r,int l,vector<vector<char>>& board, string word,int n,int m){        if(flag) return;        if(index==word.length()){            flag=true;            return;        }        for(int i=0;i<4;i++){            int rr=dic[i][0]+r;            int ll=dic[i][1]+l;            if(rr<0||rr>=n||ll<0||ll>=m||vis[rr][ll]==1||board[rr][ll]!=word[index]) continue;            vis[rr][ll]=1;            dfs(index+1,rr,ll,board,word,n,m);            vis[rr][ll]=0;        }    }    };