LeetCode: Scramble String [87]

【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.

【题意】

给定一个字符串s1, 我们可以将其分裂表示成一颗二叉树，通过调换二叉树中非空节点的左右孩子，我们可以组合得到新的字符串s2, 我们称s2是s1的scrambled string。

本题输入字符串s1和s2, 判断s2是否是s1的scrambled字符串。

【思路】

1. s1, s2长度必须相等，且拥有相同的元素，且各元素出现的次数也相同。

2. 如果s2是s1的scramble字符串，则s1一定存在一种分裂s11, s12, s2一定存在一种分裂s21,s22, 满足： isScramble(s11, s21)&&isScramble(s12, s22) || isScramble(s11, s22) && isScramble(s12, s21)   其中isScramble(x, y)表示y是x的scramble字符串。

3. 递归判断所有分裂情况。

【代码】

class Solution {
public:
    bool isValid(string s1, string s2){
        //s1和s2应该有相同种的字符，且各字符出现的次数相同
        sort(s1.begin(), s1.end());
        sort(s2.begin(), s2.end());
        if(s1.compare(s2)==0)return true;
        return false;
    }
    bool isScramble(string s1, string s2) {
        if(s1.length()!=s2.length())return false;
        if(s1.length()==0 || s2.length()==0)return false;

        if(s1.compare(s2)==0)return true;       //如果s1和s2相同，则返回true;
        if(s1.length()==1)return false;     //如果s1和s2长度为1， 且不相等，则返回false

        //考虑所有的分裂情况，然后递归判断
        for(int len=1; len<=s1.length()-1; len++){
            string s1left = s1.substr(0, len);
            string s1right = s1.substr(len, s1.length()-len);
            string s2left = s2.substr(0, len);
            string s2right = s2.substr(len, s1.length()-len);
            //在进行下一轮递归是需要先判断输入元素是否合法，如果不合法就没必要进行递归，过多的递归会使时间成本陡增
            if(isValid(s1left, s2left) && isScramble(s1left, s2left) && isScramble(s1right, s2right))return true;
            s2left = s2.substr(0, s1.length()-len);
            s2right = s2.substr(s1.length()-len, len);
            if(isValid(s1left, s2right) && isScramble(s1left, s2right) && isScramble(s1right, s2left))return true;
        }
        return false;
    }
};

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