【Leetcode】Word Search

最新推荐文章于 2025-07-29 21:57:56 发布

yyz20002008

最新推荐文章于 2025-07-29 21:57:56 发布

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文章标签： leetcode Java search String boolean

本文介绍了一个二维网格中查找指定单词的问题及其解决方案。采用深度优先搜索算法遍历网格，检查单词是否能通过相邻字母构成。文章提供了详细的Java实现代码。

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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.

Java:

http://blog.youkuaiyun.com/linhuanmars/article/details/24336987

深度优先 used[][]要看懂

public class Solution {
    
    public boolean exist(char[][] board, String word) 
    {
        if(word==null||word.length()==0) return true;
        if(board==null||board.length==0||board[0].length==0) return false;
        boolean[][] used=new boolean[board.length][board[0].length];
        for(int i=0;i<board.length;i++)
        {
            for(int j=0;j<board[0].length;j++)
            {
                if(search(board,word,0,i,j,used)) return true;
            }
        }
        return false;
    }
    
	private boolean search(char[][] board, String word, int index, int i, int j, boolean[][] used)
	{
	    if(index==word.length()) return true;
	    if(i<0||j<0||i>=board.length||j>=board[0].length||used[i][j]||board[i][j]!=word.charAt(index)) return false;
	    used[i][j]=true;
	    boolean res=search(board,word,index+1,i-1,j,used)||search(board,word,index+1,i+1,j,used)||search(board,word,index+1,i,j-1,used)||search(board,word,index+1,i,j+1,used);
	    used[i][j]=false;
	    return res;
	}
}