【Leetcode】Word Search

最新推荐文章于 2019-05-21 18:57:38 发布

原创最新推荐文章于 2019-05-21 18:57:38 发布 · 165 阅读

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本文介绍了一种使用深度优先搜索(DFS)算法在二维字符网格中查找特定单词的方法。该算法确保了同一字母单元格不会被重复使用，并且单词中的字母必须来自相邻的单元格。通过递归地检查所有可能路径来确定单词是否存在。

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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 =
[
["ABCE"],
["SFCS"],
["ADEE"]
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

思路

DFS

代码

bool dfs(vector<vector<char>>& board, string word,int x,int y,int level,vector<vector<bool> >&visited) {
         if(x<0||x>=board.size()||y<0||y>=board[0].size())return false;
         if(visited[x][y])return false;
         if(word[level]!=board[x][y])return false;
         if(level==word.size()-1)return true;
         visited[x][y]=true;

         bool flag= dfs(board,word,x+1,y,level+1,visited)
         ||dfs(board,word,x,y+1,level+1,visited)
         ||dfs(board,word,x-1,y,level+1,visited)
         ||dfs(board,word,x,y-1,level+1,visited);

         visited[x][y]=false;

         return flag;


}
bool exist(vector<vector<char>>& board, string word) {
    int m=board.size();
    int n=board[0].size();
    vector<vector<bool> >visited(m,vector<bool>(n,0));

    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
    {
        if(dfs(board,word,i,j,0,visited))return true;
    }

    return false;
}