[LeetCode]87. Scramble String

[LeetCode]87. Scramble String

题目描述

思路

对于题目中描述的扰乱字符串，可能存在多个

递归解决
当前输入的两个字符串如果相等，即返回true
否则 对i从1到len-1
判断 s1(0,i) 与 s2(0,i)或则s2(len - i, len - 1)是否满足扰乱串的条件
s1(i,len - 1) 与 s2(i, len - 1)或则s2(0, len - i)是否满足
若均满足返回true
否则返回false

代码

#include <iostream>
#include <string>
#include <vector>

using namespace std;



class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1 == s2)
            return true;
        vector<int> count(26, 0);
        for (int i = 0; i < s1.size(); ++i) {
            count[s1[i] - 'a']++;
            count[s2[i] - 'a']--;
        }

        for (int p : count) {
            if (p != 0)
                return false;
        }

        for (int i = 1; i < s1.size(); ++i) {
            if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
                return true;
            if (isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i), s2.substr(0, s1.size() - i)))
                return true;
        }
        return false;
    }
};

int main() {
    Solution s;
    cout << s.isScramble("ab", "ba") << endl;

    system("pause");
    return 0;
}