97. Interleaving String

难度: Medium

题目描述

Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that:

• s = s1 + s2 + ... + sn
• t = t1 + t2 + ... + tm
• |n - m| <= 1
• The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

 

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

 

Constraints:

• 0 <= s1.length, s2.length <= 100
• 0 <= s3.length <= 200
• s1, s2, and s3 consist of lowercase English letters.

 

Follow up: Could you solve it using only O(s2.length) additional memory space?

解题思路

代码实现

解决方案

java

var cache = make(map[string]bool)

func isInterleave(s1 string, s2 string, s3 string) bool {
	if len(s1)+len(s2) != len(s3) {
		return false
	}

	if len(s1) == 0 && len(s2) == 0 && len(s3) == 0 {
		return true
	}
	m := len(s1)
	n := len(s2)
	// t := len(s3)
	dp := make([]bool, n+1)

	 
	dp[0] = true
	for i := 0; i <= m; i++ {
		for j := 0; j <= n; j++ {
			p := i + j - 1
			if i > 0 {
				dp[j] =  (dp[j] && s1[i-1] == s3[p])
			}
			if j > 0 {
				dp[j] = dp[j] || (dp[j-1] && s2[j-1] == s3[p])
			}
		}
	}

	return dp[n]

}