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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.
class Solution { public: bool isScramble(string s1, string s2) { if(s1==s2)return true; if(s1.length()!=s2.length())return false; int length=s1.size(); int a[26]={0}; for(int i=0;i< length;i++) { a[s1[i]-‘a‘]++; a[s2[i]-‘a‘]--; } for(int i=0;i<26;i++) { if(a[i]!=0)return false; } bool f[length][length][length]; memset(f, false, sizeof(bool) * length * length * length); // vector<vector<vector<bool>>> f( length,vector<vector<bool>> ( length,vector<bool> ( length, false))); for (int k=1;k<= length;k++) { for (int i=0;i<= length-k;i++) { for (int j=0;j<= length-k;j++) { if (k==1) { f[i][j][k]=(s1[i]==s2[j]); } else { for (int kk=1;kk<k;kk++) { if ((f[i][j][kk]&&f[i+kk][j+kk][k-kk])||(f[i+kk][j][k-kk]&&f[i][j+k-kk][kk])) { f[i][j][k]=true; break; } } } } } } return f[0][0][length]; } /* bool isScramble(string s1, string s2) { // Start typing your C/C++ solution below // DO NOT write int main() function if (s1.length() != s2.length()) { return false; } int length = s1.length(); bool f[length][length][length]; memset(f, false, sizeof(bool) * length * length * length); for (int k = 1; k <= length; k++) { for (int i = 0; i <= length - k; i++) { for (int j = 0; j <= length - k; j++) { if (k == 1) { f[i][j][k] = s1[i] == s2[j]; } else { for (int l = 1; l < k; l++) { if ((f[i][j][l] && f[i + l][j + l][k - l]) || (f[i][j + k - l][l] && f[i + l][j][k - l])) { f[i][j][k] = true; break; } } } } } } return f[0][0][length]; } bool isScramble(string s1, string s2) { if(s1==s2)return true; if(s1.length()!=s2.length())return false; int len1=s1.size(), len2=s2.size(); int a[26]={0}; for(int i=0;i<len1;i++) { a[s1[i]-‘a‘]++; a[s2[i]-‘a‘]--; } for(int i=0;i<26;i++) { if(a[i]!=0)return false; } for(int i=1;i<len1;i++) { bool res1=isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i)); bool res2=isScramble(s1.substr(0,i),s2.substr(len2-i,i))&&isScramble(s1.substr(i),s2.substr(0,len2-i)); bool res=res1||res2; if(res) return res; } return false; } */ };
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原文地址：http://www.cnblogs.com/Vae98Scilence/p/4281404.html
