LeetCode97. Interleaving String—

LeetCode97. Interleaving String——动态规划

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文章目录

- 一、题目
- 二、题解

一、题目

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?

二、题解

class Solution {
public:bool isInterleave(string s1, string s2, string s3) {int n = s1.size(), m = s2.size();if(n + m != s3.size()) return false;//s1长度为i,s2长度为j,能否组成长度为i+j的s3vector<vector<bool>> dp(n+1,vector<bool>(m+1,false));dp[0][0] = true;for(int i = 1;i <= n;i++){if(s1[i-1] != s3[i-1]) break;dp[i][0] = true;}for(int j = 1;j <= m;j++){if(s2[j-1] != s3[j-1]) break;dp[0][j] = true;}for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){dp[i][j] = (s1[i-1] == s3[i+j-1] && dp[i-1][j]) || (s2[j-1] == s3[i+j-1] && dp[i][j-1]);}}return dp[n][m];}
};