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 = ; isep < s1.size(); ++ isep) { //traverse split pos
string seg11 = s1.substr(,isep);
string seg12 = s1.substr(isep);
//see if a1=a2 &&b1=b2 is ok
string seg21 = s2.substr(,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(,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。
状态转移方程：if(dp[i][j][split] && dp[i+split][j+split][k-split] || dp[i][j+k-split][split] && dp[i+split][j][k-split]) dp[i][j][k]=true;
因为在状态转移方程中k又要分割成更小的值，所以必须已知小值，k从小到大遍历。
class Solution {
public:
bool isScramble(string s1, string s2) {
int len = s1.length();
if(len==) return true;
if(s1 == s2) return true;
//初始状态
vector<vector<vector<bool>>> dp(len, vector<vector<bool>>(len, vector<bool>(len+, false) ) );
for (int i = ; i < len; ++i)
{
for (int j = ; j < len; ++j)
{
dp[i][j][] = s1[i]==s2[j];
}
}
//状态转移
for(int k = ; k <= len; k++) //从较短的子串开始分析，为了状态转方程
{
for(int s1Pointer = ; s1Pointer+k- < len; s1Pointer++)
{
for(int s2Pointer = ; s2Pointer+k- < len; s2Pointer++)
{
for(int split = ; split < k; split++) //levelSize长度的任意一种分割
{
if ((dp[s1Pointer][s2Pointer][split] && dp[s1Pointer+split][s2Pointer+split][k-split]) ||
(dp[s1Pointer][s2Pointer+k-split][split] && dp[s1Pointer+split][s2Pointer][k-split]))
{
dp[s1Pointer][s2Pointer][k] = true;
break;
};
}
}
}
}
return dp[][][len];
}
};