leetcode【87】Scramble String

问题描述

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":

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".

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".

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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

题目大意就是判断s2能够由s1的树反转得到。

解题思路：

递归的思想，每次从中间分开，查看左右能否满足条件，或者交叉满足条件。因为不知道从哪里分开，所以就需要遍历，核心代码如下：

for(int i=1; i<N; i++){
      if(isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i)))
          return true;
      if(isScramble(s1.substr(0,i), s2.substr(N-i)) && isScramble(s1.substr(i), s2.substr(0, N-i)))
          return true;
}

两个判断的意思就是s1，s2左，s1，s2右能满足条件。或者s1左s2右，s1右s2左能满足条件。

跳出递归的条件如下：

if(s1.length() != s2.length())  return false;
        int N = s1.length();
        if(N==0)    return true;
        if(N==1)    return s1==s2;

程序源码：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.length() != s2.length())  return false;
        int N = s1.length();
        if(N==0)    return true;
        if(N==1)    return s1==s2;
        string s1cpy = s1;
        string s2cpy = s2;
        sort(s1cpy.begin(), s1cpy.end());
        sort(s2cpy.begin(), s2cpy.end());
        if(s1cpy != s2cpy)  return false;
        for(int i=1; i<N; i++){
            if(isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i)))
                return true;
            if(isScramble(s1.substr(0,i), s2.substr(N-i)) && isScramble(s1.substr(i), s2.substr(0, N-i)))
                return true;
        }
        return false;
    }
};