【C++】 LeetCode 87. Scramble String

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    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.

运行结果：

解析：

题意：判断两个字符串是否能通过二叉树的左右子树交换相等

思路：

这题最简单的方法就是直接用递归，暴力求解，把s1，s2分别分成两部分，判断s1的两部分和s2的两部分是否分别可以交换相等。但是提交会超时。

超时的原因是很多比较是重复进行的，导致运算量过大。故可以采用动态规划的思想把中间结果保存起来直接调用，就可以减少重复计算了。由于此题是两个字符串，故记录结果需要用到三维数组iscmp[i][j][len]表示s1以i作为起始位，长度为len的字符串和s2以j作为起始位，长度为len的字符串是否可以交换相等。三维数组默认值均为0，则可以采用1表示可交换，-1表示不可交换，就可以不用对三位数组进行初始化了（默认值都为0）

代码：

class Solution {public:    bool findscram(string &s1, string &s2,int i1,int i2,int len,vector<vector<vector<int>>> &iscmp){        if(i1>s1.size()||i2>s2.size()){            if(i1>=s1.size()&&i2>=s2.size())return true;            else return false;        }        if(iscmp[i1][i2][len]==1) return true;        else if(iscmp[i1][i2][len]==-1) return false;                if(len==1){            if(s1[i1]==s2[i2]){                iscmp[i1][i2][len]=1;                return true;            }            else{                iscmp[i1][i2][len]=-1;                return false;            }        }        for(int i=1;i<len;i++){            if(findscram(s1,s2,i1,i2,i,iscmp)&&findscram(s1,s2,i1+i,i2+i,len-i,iscmp)){                iscmp[i1][i2][len]=1;                return true;            }            if(findscram(s1,s2,i1,i2+len-i,i,iscmp)&&findscram(s1,s2,i1+i,i2,len-i,iscmp)){                iscmp[i1][i2][len]=1;                return true;            }        }        iscmp[i1][i2][len]=-1;        return false;    }    bool isScramble(string s1, string s2) {        int len1=s1.size();        int len2=s2.size();        if(len1!=len2)return false;        if(len1==0) return true;        vector<vector<vector<int>>> iscmp(len1+1,vector<vector<int>>(len1+1,vector<int>(len1+1,0)));        return findscram(s1,s2,0,0,len1,iscmp);    }};