[LeetCode]Scramble String-优快云博客

本文链接：https://blog.youkuaiyun.com/CiaoLiang/article/details/50408385

本文介绍了一种用于判断两个字符串是否可以通过乱序操作相互转换的算法。通过递归分割字符串并进行子串比较，该算法能有效判断字符串s2是否为s1的乱序版本。文章还提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

递归剪枝。

对两个字符串可以判断排序后是否相等确认是否是scrambled的，如果不相等一定返回false,剪枝后速度加快很多。

递归过程，把s1和s2分成两个部分，sa1,sb1,sa2,sb2.

返回 ((sa1~sb1)&&(sa2~sb2))||((sa1~sb2)&&(sa2~sb1)).

(s1~s2)表示是否scrambled

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if(len1!=len2)
            return false;
        if(s1==s2)
            return true;
        int A[26] = {0};
        for(int i=0; i<len1; ++i)
            ++A[s1[i]-'a'];
        for(int i=0; i<len2; ++i)
            --A[s2[i]-'a'];
        for(int i=0; i<26; ++i){
            if(A[i]!=0)
                return false;
        } //递归剪枝，判断字符串各个字母是否相同，采用类似桶排序简化。
        for(int i=1; i<len1; ++i){
            bool ret = isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i));
            ret = ret||(isScramble(s1.substr(0,i),s2.substr(len2-i,i))&&isScramble(s1.substr(i),s2.substr(0,len2-i)));//递归解
            if(ret)
                return true;
        }
        return false;
    }
};