[LeetCode]Scramble String

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    eat / \    /  \g   r  e   at           / \          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat   /    \  rg    eat / \    /  \r   g  e   at           / \          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae   /    \  rg    tae / \    /  \r   g  ta  e       / \      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Analysis:

首先想到的是递归，简单明了，对两个string进行partition，然后比较四个字符串段。但是递归的话，这个时间复杂度比较高。然后想到能否DP，但是即使用DP的话，也要O(n^3)。想想算了，还是在递归里做些剪枝，这样就可以避免冗余计算：
对于每两个要比较的partition，统计他们字符出现次数，如果不相等返回。

网上也有动态规划的解法， 稍后研究一下

Java

public boolean isScramble(String s1, String s2) {        if(s1.length()!=s2.length()) return false;int[] frequence = new int[26];for(int i=0;i<s1.length();i++){int temp = s1.charAt(i)-'a';frequence[temp]++;}for(int i=0;i<s2.length();i++){int temp = s2.charAt(i)-'a';frequence[temp]--;}for(int i=0;i<frequence.length;i++){if(frequence[i]!=0) return false;}if(s1.length()==1 && s2.length()==1) return true;for(int i=1;i<s1.length();i++){boolean result = isScramble(s1.substring(0,i), s2.substring(0,i)) && isScramble(s1.substring(i), s2.substring(i));if(result) return true;result = (isScramble(s1.substring(0, i), s2.substring(s2.length()-i))&& isScramble(s1.substring(i), s2.substring(0, s2.length()-i)));if(result) return true;}return false;    }

c++

bool isScramble(string s1, string s2) {        if(s1.size() != s2.size()) return false;    int A[26];    memset(A,0,26*sizeof(A[0]));    for(int i=0; i<s1.size(); i++){        A[s1[i]-'a']++;    }    for(int i=0; i<s2.size(); i++){        A[s2[i]-'a']--;    }    for(int i=0; i<26; i++){        if(A[i] !=0)            return false;    }    if(s1.size() ==1 && s2.size()==1) return true;    for(int i=1; i<s1.size(); i++){        bool result = isScramble(s1.substr(0,i), s2.substr(0,i)) &&        isScramble(s1.substr(i,s1.size()-i),s2.substr(i,s1.size()-i));        result = result || (isScramble(s1.substr(0,i),s2.substr(s2.size()-i,i))&& isScramble(s1.substr(i,s1.size()-i),s2.substr(0,s1.size()-i)));        if(result) return true;    }    return false;    }