87. Scramble String

Posted 2020-10-30 小河沟大河沟

87. Scramble String

题目

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.

解析

• 【LeetCode】87. Scramble String解法及注释

题意在于判断一个字符串是否为另一个字符串“乱序”得到，这种乱序采用的方式是将一个字符串从某个位置“割开”，形成两个子串，然后对两个子串进行同样的“割开”操作，直到到达叶子节点，无法再分割。然后对非叶子节点的左右孩子节点进行交换，最后重新从左至右组合形成新的字符串，由于这个过程中存在字符位置的变化，因此，原来的字符串顺序可能会被打乱，当然也可能没有（同一个非叶子节点的左右孩子交换0次或偶数次，就无变化）。需要注意的点：

1、原字符串每次被割开的位置并不确定可能为[1,s.size()-1]，所以必然需要遍历所有可能割开的位置；

2、原字符串从第i个位置被割开（i在区间[1,s.size()-1]），形成的两个子串s.substr(0,i)和s.substr(i,s.size()-i)，如果这两个子串不全为空，则它们的母串（这里指原字符串）就是所谓的非叶子节点，这两个子串可以左右交换（按照二叉树的展开方式）；对于两个子串，可继续割裂，直到形成叶子节点。

// 87. Scramble String
class Solution_87 {
public:
    bool isScramble(string s1, string s2) {

        if (s1==s2)
        {
            return true;
        }
        if (s1.size()!=s2.size())
        {
            return false;
        }

        vector<int> hash(26,0);
        for (int i = 0; i < s1.size();i++)
        {
            hash[s1[i] - 'a']++;
            hash[s2[i] - 'a']--;
        }
        for (int i = 0; i < 26;i++)  //递归剪枝
        {
            if (hash[i]!=0)
            {
                return false;
            }
        }

        bool res = false;
        for (int i = 1; i < s1.size();i++) //遍历所有可能割开的位置, 切割的长度
        {
            res = res || (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i, s1.size() - i), s2.substr(i, s1.size() - i)));   //长度要一致
            res = res || (isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i),s2.substr(0, s1.size()-i)));
        }

        return res;

    }
};

题目来源

• 87. Scramble String