【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);
}
};