Leetcode 79. Word Search-优快云博客

本文介绍了一个算法，用于在二维字符网格中查找指定的单词。该算法通过递归地检查每个可能的路径，确保同一单元格不会被重复使用，并且只在水平或垂直相邻的方向上移动。提供了完整的Java实现代码。

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 boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0 || board[0].length == 0) return false;
        int row = board.length, col = board[0].length;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (word.charAt(0) != board[i][j]) continue;
                if (exist(board, i, j, word, 0)) return true;
            }
        }
        return false;
    }

    private boolean exist(char[][] board, int x, int y, String word, int start) {
        if(start == word.length()) return true;
        if(x < 0 || x >= board.length || y < 0 || y >= board[0].length || board[x][y] != word.charAt(start) || board[x][y] == '#') return false;
        if (board[x][y] == word.charAt(start)) {
            char c = board[x][y];
            board[x][y] = '#';
            boolean res = exist(board, x + 1, y, word, start + 1) || 
                          exist(board, x - 1, y, word, start + 1) ||
                          exist(board, x, y + 1, word, start + 1) || 
                          exist(board, x, y - 1, word, start + 1);
            board[x][y] = c;
            return res;
        }
        return false;
    }