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.
地址：https://oj.leetcode.com/problems/scramble-string/
算法：如果字串S1[0~i]和S2[0~i]以及S1[i+1~n]和S2[i+1~n]都满足scrambled条件，那么S1和S2也满足；如果字串S1[0~i]和S2[n-i~n]以及S1[i+1~n]和S2[0~n-i-1]满足scrambled条件，那么S1和S2也满足，其中i=1...n。注意，我们在递归调用之前，用函数isReasonString先判断一下是否有可能满足scrambled条件，这样可以减少递归的次数。代码：
1 class Solution {
2 public:
3 bool isScramble(string s1, string s2) {
4 if(s1 == s2){
5 return true;
6 }
7 if(s1.size() == 1){
8 return false;
9 }
10 int len = s1.size();
11 for(int i = 1; i < len; ++i){
12 string s11 = s1.substr(0,i);
13 string s21 = s2.substr(0,i);
14 string s12 = s1.substr(i,len-i);
15 string s22 = s2.substr(i,len-i);
16 if(isReasonString(s11,s21) && isReasonString(s12,s22) && isScramble(s11,s21) && isScramble(s12,s22)){
17 return true;
18 }
19 s21 = s2.substr(len-i,i);
20 s22 = s2.substr(0,len-i);
21 if(isReasonString(s11,s21) && isReasonString(s12,s22) && isScramble(s11,s21) && isScramble(s12,s22)){
22 return true;
23 }
24 }
25 return false;
26 }
27 bool isReasonString(string s1, string s2){
28 sort(s1.begin(),s1.end());
29 sort(s2.begin(),s2.end());
30 return s1 == s2;
31 }
32 };