题目： Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 ="great": great/ \gr ea

3D dp问题 class Solution {public:bool isScramble(string s1, string s2) {// Start typing your C/C++ solution below// DO NOT write int main() functionint N = s1.size();if(N != s2.size()) return false;

题意： Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great /    \ gr

87. Scramble String 暴力递归算法： class Solution {public boolean isScramble(String s1, String s2) {if(s1.length()!=s2.length())return false;if (s1.equals(s2)) return true;int[] record=new int[26];for(int i

题目描述： Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great":     great    /

题解 符号： S_(i-j):表示字符串s下标从i到j的字串。 思路 其实没什么规律，就是暴力枚举交换轴，然后每次有交换与不交换两种情况，递归判断是否可行。唯一剪枝就是假如S1_(i,j)=S2_(k,l)，则他们所包含的字母的集合是相同的，如果不同，则不用再继续递归下去。 代码 class Solution {public:int *sum1,*sum2;bool dfs(int l1

题目：搅乱字符串 Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great/ \gr

[[Leetcode]87. Scramble String](https://leetcode.com/problems/scramble-string/) - 本题难度: Hard - Topic: divide and conquere # Description Given a string s1, we may represent it as a binary tree by par

题意： 判断两个字符串是否互为Scramble字符串，而互为Scramble字符串的定义： 字符串看作是父节点，从字符串某一处切开，生成的两个子串分别是父串的左右子树，再对切开生成的两个子串继续切开，直到无法再切，此时生成为一棵二叉树。对二叉树的任一子树可任意交换其左右分支，如果S1可以通过交换变成S2，则S1,S2互为Scramble字符串。 思路： 对于分割后的子串，应有IsScrambl

参考后， 没做出来 class Solution {public:bool isScramble(string s1, string s2) {vector<int> v(127);size_t n = s1.size();if (n == 1) return s1[0] == s2[0];for (size_t i = 0; i != n; ++i) {++v[s1[i]];--v[s2[i