[Leetcode]Word Search

2024年6月8日 190次阅读

Word Search
Total Accepted: 58219 Total Submissions: 270800 Difficulty: Medium

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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用DFS做的，中间调试了一会，不过感觉代码还算得上干净。

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size();
        if(!m)  return false;
        int n = board[0].size();
        bool res = false;
        vector<vector<bool> > used(m,vector<bool>(n,false));
        for(int i = 0;i <= m - 1;++ i){
            for(int j = 0;j <= n - 1;++ j){
                if(used[i][j] == false && board[i][j] == word[0]){
                    used[i][j] = true;
                    res = dfs(board,i,j,word,1,used);
                    if(res == true) return true;
                    used[i][j] = false;
                }
            }
        }
        return false;
    }
private:
    bool dfs(const vector<vector<char> >& board,int i,int j,const string& word,int index,vector<vector<bool> >& used){
        bool res = false;
        int m = board.size();
        int n = board[0].size();
        if(index == word.size()){
            res = true;
            return res;
        }
        if(i - 1 >= 0 && !used[i - 1][j] && word[index] == board[i - 1][j]){
            used[i - 1][j] = true;
            res = dfs(board,i - 1,j,word,index + 1,used);
            used[i - 1][j] = false;
            if(res) return true;
        }
        if(i + 1 <= m - 1 && !used[i + 1][j] && word[index] == board[i + 1][j]){
            used[i + 1][j] = true;
            res = dfs(board,i + 1,j,word,index + 1,used);
            used[i + 1][j] = false;
            if(res) return true;
        }
        if(j - 1 >= 0 && !used[i][j - 1] && word[index] == board[i][j - 1]){
            used[i][j - 1] = true;
            res = dfs(board,i,j - 1,word,index + 1,used);
            used[i][j - 1] = false;
            if(res) return true;
        }
        if(j + 1 <= n - 1 && !used[i][j + 1] && word[index] == board[i][j + 1]){
            used[i][j + 1] = true;
            res = dfs(board,i,j + 1,word,index + 1,used);
            used[i][j + 1] = false;
            if(res) return true;
        }
        return false;
    }
};

看这那四个判断语句好累赘，其实可以用循环替代，因为如果对角线也算的话就会有8个循环，这是很丑陋的。可以用数组存储方向，然后在循环里if进行递归。
有时间美化下吧

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