Word Search

描述

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 neighbouring. The same letter cell may not be used more than once.

For example, Given board =

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

word = "ABCCED", -> returns true,

word = "SEE", -> returns true,

word = "ABCB", -> returns false.

分析

无。

代码

// Word Search
// 深搜，递归
// 时间复杂度O(n^2*m^2)，空间复杂度O(n^2)
public class Solution {
    public boolean exist(char[][] board, String word) {
        final int m = board.length;
        final int n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (dfs(board, word, 0, i, j, visited))
                    return true;
        return false;
    }
    private static boolean dfs(char[][] board, String word,
                    int index, int x, int y, boolean[][] visited) {
        if (index == word.length())
            return true; // 收敛条件

        if (x < 0 || y < 0 || x >= board.length || y >= board[0].length)
            return false;  // 越界，终止条件

        if (visited[x][y]) return false; // 已经访问过，剪枝

        if (board[x][y] != word.charAt(index)) return false; // 不相等，剪枝

        visited[x][y] = true;
        boolean ret = dfs(board, word, index + 1, x - 1, y, visited) || // 上
                dfs(board, word, index + 1, x + 1, y, visited)    || // 下
                dfs(board, word, index + 1, x, y - 1, visited)    || // 左
                dfs(board, word, index + 1, x, y + 1, visited);      // 右
        visited[x][y] = false;
        return ret;
    }
}