87. Scramble String (String; DP)

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.

思路：对付复杂问题的方法是从简单的特例来思考。简单情况：

如果字符串长度为1，那么必须两个字符串完全相同；

如果字符串长度为2，例如s1='ab'，则s2='ab'或s2='ba'才行

如果字符串任意长度，那么可以把s1分为a1, b1两部分，s2分为a2,b2两部分。需要满足：((a1=a2)&&(b1=b2)) || ((a1=b2)&&(a2=b1)) =>可用递归

class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1 == s2) return true;
for(int isep = 1; isep < s1.size(); ++ isep) { //traverse split pos
string seg11 = s1.substr(0,isep);
string seg12 = s1.substr(isep);

//see if a1=a2 &&b1=b2 is ok
string seg21 = s2.substr(0,isep);
string seg22 = s2.substr(isep);
if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true;

//see if a1=b2 &&a2=b1 is ok
seg21 = s2.substr(s2.size() - isep); //从后截取isep长度
seg22 = s2.substr(0,s2.size() - isep);
if(isScramble(seg11,seg21) && isScramble(seg12,seg22)) return true;
}
return false;
}
};

Result: Time Limit Exceeded

思路II: 动态规划。三维状态dp[i][j][k]，前两维分别表示s1和s2的下标起始位置，k表示子串的长度。dp[i][j][k]=true表示s1(i, i+k-1)和s2(j, j+k-1)是scramble。

class Solution {
public:
bool isScramble(string s1, string s2) {
int len = s1.length();
if(len==0) return true;
if(s1 == s2) return true;

//初始状态
vector<vector<vector<bool>>> dp(len, vector<vector<bool>>(len, vector<bool>(len+1, false) ) );
for (int i = 0; i < len; ++i)
{
for (int j = 0; j < len; ++j)
{
dp[i][j][1] = s1[i]==s2[j];
}
}

//状态转移
for(int levelSize = 2; levelSize <= len; levelSize++) //从较短的子串开始分析，为了状态转方程
{
for(int s1Pointer = 0; s1Pointer+levelSize-1 < len; s1Pointer++)
{
for(int s2Pointer = 0; s2Pointer+levelSize-1 < len; s2Pointer++)
{
for(int split = 1; split < levelSize; split++) //levelSize长度的任意一种分割
{
if ((dp[s1Pointer][s2Pointer][split] && dp[s1Pointer+split][s2Pointer+split][levelSize-split]) ||
(dp[s1Pointer][s2Pointer+levelSize-split][split] && dp[s1Pointer+split][s2Pointer][levelSize-split]))
{
dp[s1Pointer][s2Pointer][levelSize] = true;
break;
};
}
}
}
}
return dp[0][0][len];
}
};