Word Search

ncst

于 2018-04-03 22:14:55 发布

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分类专栏： leetcode

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本文介绍了一个二维网格中查找特定单词的问题及其解决方案。采用深度优先搜索（DFS）算法遍历网格，寻找与目标单词匹配的路径。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.

Have you met this question in a real interview?

Yes

Example

Given board =

[
  "ABCE",
  "SFCS",
  "ADEE"
]

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

可以直接使用DFS进行处理，处理后的结果直接返回。

java

public class Solution {
    /**
     * @param board: A list of lists of character
     * @param word: A string
     * @return: A boolean
     */
    public boolean exist(char[][] board, String word) {
        // write your code here
        if (board == null || board.length == 0 || board[0] == null || 
            board[0].length == 0) {
            return false;        
        }
        if (word == null || word.length() == 0) {
            return true;
        }
        boolean res = false;
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == word.charAt(0)) {
                    if (find(board, i, j, 0, word)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
    private boolean find(char[][] board, int i, int j, int index, 
                         String word) {
        if (index >= word.length()) {
            return true;
        }
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length ||
            word.charAt(index) != board[i][j]) {
            return false;        
        }
        board[i][j] = '#';
        boolean res = find(board, i + 1, j, index + 1, word) || 
              find(board, i - 1, j, index + 1, word) ||
              find(board, i, j + 1, index + 1, word) ||
              find(board, i, j - 1, index + 1, word);
        board[i][j] = word.charAt(index);
        return res;
    }
}