97. 交错字符串 @Python @Go

原题

https://leetcode.cn/problems/interleaving-string/description/

思路

动态规划

复杂度

时间：O(len(s1) * len(s2))
空间：O(len(s1) * len(s2))

Python代码

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        l1 = len(s1)
        l2 = len(s2)
        l3 = len(s3)
        if l1 + l2 != l3:
            return False
        dp = [[False] * (l2+1) for i in range((l1+1))]
        dp[0][0] = True 
        for i in range(1, l1+1):
            dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1] 
        for j in range(1, l2+1):
            dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
        for i in range(1, l1+1):
            for j in range(1, l2+1):
                dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i+j-1]) or (dp[i][j-1] and s2[j-1] == s3[i+j-1])

        return dp[-1][-1]        

Go代码

func isInterleave(s1 string, s2 string, s3 string) bool {
	l1, l2, l3 := len(s1), len(s2), len(s3)
	if l1+l2 != l3 {
		return false
	}
	dp := make([][]bool, l1+1)
	for i := 0; i < l1+1; i++ {
		dp[i] = make([]bool, l2+1)
	}
    // 初始化
    dp[0][0] = true 
	for i := 1; i < l1+1; i++ {
		dp[i][0] = dp[i-1][0] && s1[i-1] == s3[i-1]
	}
	for j := 1; j < l2+1; j++ {
		dp[0][j] = dp[0][j-1] && s2[j-1] == s3[j-1]
	}
	for i := 1; i < l1+1; i++ {
		for j := 1; j < l2+1; j++ {
			dp[i][j] = (dp[i-1][j] && s1[i-1] == s3[i+j-1]) || (dp[i][j-1] && s2[j-1] == s3[i+j-1])
		}
	}
	return dp[l1][l2]
}