LeetCode-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.
class Solution { public: bool isS(string& s1,string& s2,int begin1,int begin2,int end1,int end2) { int length1=end1-begin1+1; int length2=end2-begin2+1; if(length1!=length2)return false; if(length1==1){ if(s1[begin1]==s2[begin2])return true; else return false; } else{ vector<int> count; count.resize(26); for(int i=0;i<26;i++)count[i]=0; for(int i=begin1;i<=end1;i++){ count[s1[i]-'a']++; } for(int i=begin2;i<=end2;i++){ count[s2[i]-'a']--; } for(int i=0;i<26;i++){ if(count[i]!=0){ return false; } } for(int i=1;i<length1;i++){ bool b1,b2,b3,b4; int j=length1-i; b1=isS(s1,s2,begin1,begin2,begin1+i-1,begin2+i-1); b2=isS(s1,s2,begin1+i,begin2+i,end1,end2); b3= isS(s1,s2,begin1,end2-i+1,begin1+i-1,end2); b4=isS(s1,s2,begin1+i,begin2,end1,begin2+j-1); if( b1&&b2 || b3&&b4 ) return true; } return false; } } bool isScramble(string s1, string s2) { // Start typing your C/C++ solution below // DO NOT write int main() function return isS(s1,s2,0,0,s1.length()-1,s2.length()-1); } };