[LeetCode] 79. 单词搜索

题目链接 : https://leetcode-cn.com/problems/word-search/

题目描述:

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

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:

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

给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.

思路:

回溯算法,直接看代码,很容易理解

代码:

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        row = len(board)
        col = len(board[0])

        def helper(i, j, k, visited):
            #print(i,j, k,visited)
            if k == len(word):
                return True
            for x, y in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                tmp_i = x + i
                tmp_j = y + j
                if 0 <= tmp_i < row and 0 <= tmp_j < col and (tmp_i, tmp_j) not in visited 
                and board[tmp_i][tmp_j] == word[k]:
                    visited.add((tmp_i, tmp_j))
                    if helper(tmp_i, tmp_j, k+1, visited):
                        return True
                    visited.remove((tmp_i, tmp_j)) # 回溯
            return False
        
        for i in range(row):
            for j in range(col):
                if board[i][j] == word[0] and helper(i, j, 1,{(i, j)}) :
                        return True
        return False

java

class Solution {
     public boolean exist(char[][] board, String word) {
        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 (word.charAt(0) == board[i][j] && backtrack(i, j, 0, word, visited, board)) return true;
            }
        }
        return false;

    }

    private boolean backtrack(int i, int j, int idx, String word, boolean[][] visited, char[][] board) {
        if (idx == word.length()) return true;
        if (i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word.charAt(idx) || visited[i][j])
            return false;
        visited[i][j] = true;
        if (backtrack(i + 1, j, idx + 1, word, visited, board) || backtrack(i - 1, j, idx + 1, word, visited, board) || backtrack(i, j + 1, idx + 1, word, visited, board) || backtrack(i, j - 1, idx + 1, word, visited, board))
            return true;
        visited[i][j] = false; // 回溯
        return false;
    }
}