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.
思路:使用递归的方法,判断s1和s2的子串是否分别可以进行重组,如果都可以,则返回true。否则,返回false。
代码:
class Solution {
public:
bool isScramble(string s1, string s2) {
string s1_temp = s1, s2_temp = s2;
if(s1==s2)
return true;
if(s1.length()!=s2.length())
return false;
sort(s1_temp.begin(),s1_temp.end());
sort(s2_temp.begin(),s2_temp.end());
if(s1_temp!=s2_temp)
return false;
int size = s1.length();
for(int i=1;i<size;i++)
{
string s1left = s1.substr(0,i);
string s2left = s2.substr(0,i);
string s1right = s1.substr(i);
string s2right = s2.substr(i);
if(isScramble(s1left,s2left)&&isScramble(s1right,s2right))
return true;
s2left = s2.substr(0, size-i);
s2right = s2.substr(size-i);
if(isScramble(s1left,s2right)&&isScramble(s1right,s2left))
return true;
}
return false;
}
};【Leetcode】Scramble String,布布扣,bubuko.com
原文:http://blog.csdn.net/hjiam2/article/details/23690077