LeetCode 79. Word Search（单词查找）

最新推荐文章于 2020-04-18 12:04:08 发布

原创最新推荐文章于 2020-04-18 12:04:08 发布 · 1.2k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

深度优先搜索同时被 3 个专栏收录

50 篇文章

订阅专栏

字符串

30 篇文章

订阅专栏

重复

28 篇文章

订阅专栏

本文提供了一种使用深度优先搜索解决LeetCode单词搜索问题的方法。通过递归地检查每个可能的方向来查找给定单词是否存在于二维字符网格中。

原题网址：https://leetcode.com/problems/word-search/

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 .

方法：深度优先搜索。

public class Solution {
    private boolean find(char[][] board, boolean[][] used, char[] wa, int pos, int row, int col) {
        if (pos == wa.length) return true;
        if (row < 0 || row >= board.length || col < 0 || col >= board[row].length) return false;
        if (used[row][col]) return false;
        if (wa[pos] != board[row][col]) return false;
        used[row][col] = true;
        if (find(board, used, wa, pos+1, row, col-1)) return true;
        if (find(board, used, wa, pos+1, row, col+1)) return true;
        if (find(board, used, wa, pos+1, row-1, col)) return true;
        if (find(board, used, wa, pos+1, row+1, col)) return true;
        used[row][col] = false;
        return false;
    }
    public boolean exist(char[][] board, String word) {
        boolean[][] used = new boolean[board.length][board[0].length];
        char[] wa = word.toCharArray();
        for(int i=0; i<board.length; i++) {
            for(int j=0; j<board[i].length; j++) {
                if (find(board, used, wa, 0, i, j)) return true;
            }
        }
        return false;
    }
}