leetcode 761. Special Binary String

最新推荐文章于 2021-10-29 20:56:20 发布

景行cmy

最新推荐文章于 2021-10-29 20:56:20 发布

阅读量205

点赞数

分类专栏： leetcode 递归字符串

本文链接：https://blog.youkuaiyun.com/cmy203/article/details/99886140

版权

leetcode 同时被 3 个专栏收录

155 篇文章

订阅专栏

字符串

18 篇文章

订阅专栏

递归

6 篇文章

订阅专栏

本文探讨了特殊二进制字符串的性质与操作，提出了通过交换连续的特殊子字符串来生成字典序最大字符串的算法。该算法首先解析输入字符串为多个特殊子字符串，然后按字典序逆序排列并重新组合，最终返回重构后的最大字符串。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Special binary strings are binary strings with the following two properties:

The number of 0's is equal to the number of 1's.
Every prefix of the binary string has at least as many 1's as 0's.

Given a special string S, a move consists of choosing two consecutive, non-empty, special substrings of S, and swapping them. (Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string.)

At the end of any number of moves, what is the lexicographically largest resulting string possible?

Example 1:

Input: S = "11011000"
Output: "11100100"
Explanation:
The strings "10" [occuring at S[1]] and "1100" [at S[3]] are swapped.
This is the lexicographically largest string possible after some number of swaps.

Note:

S has length at most 50.
S is guaranteed to be a specialbinary string as defined above.

class Solution {
public:
    string makeLargestSpecial(string S) 
    {
        set<string> strs ;
        string res ;
        for(int k = 2 ; k <= S.size() ; k++)
        {
            for(int i = 0 ; i <= S.size() - k ; ++i)
            {
                string prev = S.substr(0 , i) , suff = S.substr(i + k) , medium = S.substr(i , k) ;
                makeLargestSpecial(prev , medium , suff , strs) ;
            }
        }
        return res = *next(strs.begin() , strs.size() - 1) ;
    }
    
    void makeLargestSpecial(string prev , string s , string suff , set<string>& strs)
    {
        for(int i = 1 ; i < s.size() ; ++i)
        {
            string medium = s.substr(i) + s.substr(0 , i) ;
            strs.insert(prev + medium + suff) ;
        }
    }
};

class Solution {
public:
    string makeLargestSpecial(string S) {
        int cnt = 0, i = 0;
        vector<string> v;
        string res = "";
        for (int j = 0; j < S.size(); ++j) {
            cnt += (S[j] == '1') ? 1 : -1;
            if (cnt == 0) {
                v.push_back('1' + makeLargestSpecial(S.substr(i + 1, j - i - 1)) + '0');
                i = j + 1;
            }
        }
        sort(v.begin(), v.end(), greater<string>());
        for (int i = 0; i < v.size(); ++i) res += v[i];
        return res;
    }
};