79. Word Search

最新推荐文章于 2024-02-12 20:14:25 发布

BigFatSheep

最新推荐文章于 2024-02-12 20:14:25 发布

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分类专栏： Leetcode Medium

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/84196097

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本文介绍了一种使用深度优先搜索(DFS)算法在字符的二维数组中查找特定单词的方法。通过迭代地从每个位置开始进行DFS，如果单词长度为0则返回真，如果数组为空则返回假。在DFS过程中，检查边界条件，标记已访问的单元格，并在四个方向上递归调用DFS。

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Description

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.

Example:

board =
[
[‘A’,‘B’,‘C’,‘E’],
[‘S’,‘F’,‘C’,‘S’],
[‘A’,‘D’,‘E’,‘E’]
]

Given word = “ABCCED”, return true.
Given word = “SEE”, return true.
Given word = “ABCB”, return false.
Problem URL

Solution

给一个字符的二维数组，搜索数组中是否存在给定的单词。

We could simply run a dfs to solve this problem. If word is with length 0, return true; If board is empty, return false. Then iteratively start dfs in each position.

In dfs, if out of bound or visited or word is different, retrun false, if match, return true. Set visited to true and count ++. Then recursively call dfs in four direction, remember reset visited to false for backtracking.

Code

class Solution {
    private int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; 
    public boolean exist(char[][] board, String word) {
        if (word.length() == 0){
            return true;
        }
        if (board.length == 0 || board[0].length == 0){
            return false;
        }
        boolean[][] visited = 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 (dfs(board, word, i, j , 0, visited)){
                    return true;
                }
            }
        }
        return false;
    }
    
    private boolean dfs(char[][] board, String word, int i, int j, int count, boolean[][] visited){
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || visited[i][j] || board[i][j] != word.charAt(count)){
            return false;
        }
        if (count == word.length() - 1){
            return true;
        }
        visited[i][j] = true;
        count++;
        for (int[] dir : dirs){
            int x = i + dir[0];
            int y = j + dir[1];
            if (dfs(board, word, x, y, count, visited)){
                return true;
            }
        }
        visited[i][j] = false;
        return false;
    }
}