LeetCode 79. Word Search

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本文介绍了一种使用回溯加深度优先搜索算法解决二维网格中查找指定单词的问题。通过递归方式遍历网格，检查单词是否能由相邻字母构成，同一字母单元格不可重复使用。

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;
        }
    }
    
};