Leetcode 79 Word Search

最新推荐文章于 2020-06-06 22:45:26 发布

pessis1

最新推荐文章于 2020-06-06 22:45:26 发布

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分类专栏： Backtracking Leetcode

本文链接：https://blog.youkuaiyun.com/pessis1/article/details/73521997

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本文介绍了一种算法，用于在一个二维字符网格中查找指定的单词。该算法通过深度优先搜索逐个匹配单词中的字母，并确保同一字母单元格不会被重复使用。文章提供了完整的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 =

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

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

public class Solution {
    public boolean exist(char[][] board, String word) {
        char[] wordlist = word.toCharArray();
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(helper(board, i ,j , wordlist, 0 )){
                    return true;
                }
            }
        }
        return false;
    }
    
    
    
    private boolean helper(char[][] board, int i, int j , char[] word, int index){
        if(index == word.length){
            return true;
        }
        if(i < 0 || j < 0 || i > board.length - 1 || j > board[0].length - 1 || index > word.length - 1){
            return false;
        }
        
        if(board[i][j] != word[index]){
            return false;
        }
        
        board[i][j] ^= 256;//这里参考了网上的思路 很神奇 把char变成无效的
        boolean result = helper(board, i + 1, j , word, index + 1) || 
        helper(board, i , j + 1, word, index + 1)||
        helper(board, i - 1, j , word, index + 1)||
        helper(board, i, j - 1 , word, index + 1);
        board[i][j] ^= 256;
        return result;
    }
}