一天一道LeetCode

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（一）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.

（二）解题

题目大意：判断两个字符串是不是Scramble String，题目有点长，需要读好久！！！

举个简单的例子说明Scramble String，如ab和ba，那么ab拆分成a和b，ba拆分成b和a，这样就代表为Scramble String。

那么s1和s2是否为Scramble String，就需要将s1和s2拆分成两部分s11和s12，s21和s22，需要判断这两部分是否为Scramble String。

可以分成两个部分：（s11和s21，s12和s22）以及（s11和s22，s12和s21）。

这样一来代码就比较好写了！

很明显的动态规划问题。但是本人在做题过程中的优化还是没有做好，导致超时。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        if (len1 != len2) return false;//长度不一样
        if (s1 == s2) return true;//子串相等

        //判断两个子串包含的字符是否相同
        //最开始没加这一步，导致一直超时
        vector<int> count(26,0);
        for(int i = 0 ; i < len1 ; i++)
        {
            count[s1[i]-'a']++;
            count[s2[i]-'a']--;
        }
        for(int i = 0 ; i < 26 ; i++)
        {
            if(count[i]!=0) return false;
        }
        //递归
        for (int i = 1; i < len1; i++)
        {
            string subs1_1 = s1.substr(0, i);
            string subs1_2 = s1.substr(i);
            string subs2_1 = s2.substr(0, i);
            string subs2_2 = s2.substr(i);
            bool is1 = (subs1_1==subs2_1?true:isScramble(subs1_1, subs2_1)) && (subs1_2==subs2_2?true:isScramble(subs1_2, subs2_2));//这里也对递归进行了优化，相等则不进行递归
            subs2_1 = s2.substr(len2-i, i);
            subs2_2 = s2.substr(0,len2-i);
            bool is2 = (subs1_1==subs2_1?true:isScramble(subs1_1, subs2_1)) && (subs1_2==subs2_2?true:isScramble(subs1_2, subs2_2));
            if (is1 || is2) return true;//两部分任一部分为Scramble String则s1和s2就为Scramble String
        }
        return false;
    }
};