87. Scramble String

題目

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解法及註釋

題意在於判斷一個字符串是否為另一個字符串“亂序”得到，這種亂序采用的方式是將一個字符串從某個位置“割開”，形成兩個子串，然後對兩個子串進行同樣的“割開”操作，直到到達葉子節點，無法再分割。然後對非葉子節點的左右孩子節點進行交換，最後重新從左至右組合形成新的字符串，由於這個過程中存在字符位置的變化，因此，原來的字符串順序可能會被打亂，當然也可能沒有（同一個非葉子節點的左右孩子交換0次或偶數次，就無變化）。需要註意的點：

1、原字符串每次被割開的位置並不確定可能為[1,s.size()-1]，所以必然需要遍歷所有可能割開的位置；

2、原字符串從第i個位置被割開（i在區間[1,s.size()-1]），形成的兩個子串s.substr(0,i)和s.substr(i,s.size()-i)，如果這兩個子串不全為空，則它們的母串（這裏指原字符串）就是所謂的非葉子節點，這兩個子串可以左右交換（按照二叉樹的展開方式）；對於兩個子串，可繼續割裂，直到形成葉子節點。
 

// 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++)  //遞歸剪枝
        {
            if (hash[i]!=0)
            {
                return false;
            }
        }

        bool res = false;
        for (int i = 1; i < s1.size();i++) //遍歷所有可能割開的位置, 切割的長度
        {
            res = res || (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i, s1.size() - i), s2.substr(i, s1.size() - i)));   //長度要一致
            res = res || (isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i),s2.substr(0, s1.size()-i)));
        }

        return res;

    }
};

題目來源

• 87. Scramble String

87. Scramble String