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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 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 };
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原文地址：http://www.cnblogs.com/reachteam/p/4234925.html
