LeetCode 87. Scramble String

题目链接

87. 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.

解：

"rgtae"是"great"的一个scrambled string，"eat"也是"tae"的一个scrambled string，可见由great到rgtae，根结点左右子树未交换，从eat到tae，根结点左右子树发生交换，从这里分成2种情况：

判断两个string是不是互为scrambled string时：

1）s1前面长度i子串和s2前面长度i子串互为scrambled string && s1剩下部分串和s2剩下部分串互为scrambled string

2）s1前面长度i子串和s2后面长度i子串互为scrambled string && s1剩下部分串和s2剩下部分串互为scrambled string

遍历子串，递归，即得到结果，为提高效率，实现sameLetters函数，用以判断2个串的字符是否完全相同，毕竟字符完全相同才可能互为scrambled string，先对子串sameLetters检查，若满足再进行子串isScramble的检查，代码：

class Solution {
public:
    //2个string是否含有完全相同的字母
    bool sameLetters(string a, string b) {
        int len = a.size();
        int aa[26] = {0};
        int bb[26] = {0};
        for (int i = 0; i < len; i++) {
            aa[a[i]-'a']++;
            bb[b[i]-'a']++;
        }
        for (int i = 0; i < 26; i++) 
            if (aa[i] != bb[i]) return false;
        return true;
    }
    bool isScramble(string s1, string s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if (len1 != len2) return false;
        if (s1 == s2) return true;
        for (int i = 1; i < len1; i++) {
            string sub1Head = s1.substr(0, i);
            string sub1Tail = s1.substr(len1-i, i);
            string sub2Head = s2.substr(0, i);
            if (sameLetters(sub1Head, sub2Head))
                if (isScramble(sub1Head, sub2Head) && isScramble(s1.substr(i, len1-i), s2.substr(i, len1-i))) return true;
            if (sameLetters(sub2Head, sub1Tail)) 
                if (isScramble(sub2Head, sub1Tail) && isScramble(s2.substr(i, len1-i), s1.substr(0, len1-i))) return true;
        }
        return false;
    }
};