79. 单词搜索 @Python @Go

原题

https://leetcode.cn/problems/word-search/description/

思路

回溯法

复杂度

时间：O(m * n * len(word))
空间：O(1)

Python代码

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def backtrack(i, j, k):
            if board[i][j] != word[k]:
                return False 
            if k == len(word)-1:
                return True 
            ch = board[i][j]
            board[i][j] = '#'
            for di, dj in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                if 0 <= i + di < m and 0 <= j + dj < n:
                    if backtrack(i+di, j+dj, k+1):
                        return True 
            # 还原
            board[i][j] = ch
            return False 


        m = len(board)
        n = len(board[0])
      
        for i in range(m):
            for j in range(n):
                if backtrack(i, j, 0):
                    return True 
        return False 
        

Go代码

func exist(board [][]byte, word string) bool {
	m, n := len(board), len(board[0])
	// 匿名函数
	var backtrack func(int, int, int) bool
	backtrack = func(i int, j int, k int) bool {
		if board[i][j] != word[k] {
			return false
		}
		if k == len(word)-1 {
			return true
		}
		ch := board[i][j]
		board[i][j] = '#'
		directions := [][]int{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}
		for _, direction := range directions {
			di, dj := direction[0], direction[1]
			if 0 <= i+di && i+di < m && 0 <= j+dj && j+dj < n {
				if backtrack(i+di, j+dj, k+1) {
					return true
				}
			}
		}
		// 还原
		board[i][j] = ch
		return false
	}

	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if backtrack(i, j, 0) {
				return true
			}
		}
	}
	return false
}