【递归 &回溯】LeetCode-79. 单词搜索

79. 单词搜索。

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

提示：

m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board 和 word 仅由大小写英文字母组成

算法分析

解题思路

DFS

从单词矩阵中枚举每个单词的起点，从该起点出发往四周dfs搜索目标单词，并记录当前枚举到第几个单词，若当前搜索到的位置(i,j)的元素恰好是word单词第depth个字符，则继续dfs搜索，直到depth到最后一个字符则表示有了符合的方案，返回true

注意：搜索过的位置继续搜索下一层时，需要对当前位置进行标识，表示已经搜索

class Solution {
    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                if (dfs(board, words, 0, i, j)) return true;
            }
        }
        return false;
    }
    
    int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1}; // 左上右下
    private boolean dfs(char[][] board, char[] words, int u, int x, int y) {
        if (board[x][y] != words[u]) return false;
        if (u == words.length - 1) return true;

        char t = board[x][y];
        board[x][y] = '.';
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= board.length || b < 0 || b >= board[0].length || board[a][b] == '.') continue;
            if (dfs(board, words, u + 1, a, b)) return true;
        }
        board[x][y] = t;
        return false;
    }
}

复杂性分析

时间复杂度：O(mn * 3^l)
空间复杂度：O(m+n)