[LeetCode] Scramble String(树的问题最易用递归)
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.size()==0 || s2.size()==0) return false; if(s1 == s2) return true; string a1 = s1,a2 = s2; sort(a1.begin(),a1.end()); sort(a2.begin(),a2.end()); if(a1!= a2) return false; int len = s1.size(); for(int n = 1;n < len;n++){ if(isScramble(s1.substr(0,n),s2.substr(0,n)) && isScramble(s1.substr(n,len-n),s2.substr(n,len-n))) return true; if(isScramble(s1.substr(0,n),s2.substr(len-n,n)) && isScramble(s1.substr(n,len-n),s2.substr(0,len-n))) return true; }//end for return false; }//end func };