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.
1. 如果两个substring相等的话，则为true
2. 如果两个substring中间某一个点，左边的substrings为scramble string， 同时右边的substrings也为scramble string，则为true
3. 如果两个substring中间某一个点，s1左边的substring和s2右边的substring为scramble string, 同时s1右边substring和s2左边的substring也为scramble string，则为true
1 class Solution {
2 public:
3 bool isScramble(string s1, string s2) {
4
5 int n=s1.length();
6 vector<vector<vector<bool>>> dp(n,vector<vector<bool>>(n,vector<bool>(n+1)));
7 //dp[i][j][k] represent whether s1[i,i+1,...,i+k-1] and s2[j,j+1,...,j+k-1] is scramble
8
9
10 for(int i=n-1;i>=0;i--)
11 {
12 for(int j=n-1;j>=0;j--)
13 {
14 for(int k=1;k<=n-max(i,j);k++)
15 {
16 if(s1.substr(i,k)==s2.substr(j,k))
17 {
18 dp[i][j][k]=true;
19 }
20 else
21 {
22 for(int l=1;l<k;l++)
23 {
24 if(dp[i][j][l]&&dp[i+l][j+l][k-l]||dp[i][j+k-l][l]&&dp[i+l][j][k-l])
25 {
26 dp[i][j][k]=true;
27 break;
28 }
29 }
30
31 }
32 }
33 }
34
35 }
36 return dp[0][0][n];
37 }
38 };