一、题目

You are given a binary string s that contains at least one ‘1’.

You have to rearrange the bits in such a way that the resulting binary number is the maximum odd binary number that can be created from this combination.

Return a string representing the maximum odd binary number that can be created from the given combination.

Note that the resulting string can have leading zeros.

Example 1:

Input: s = “010”
Output: “001”
Explanation: Because there is just one ‘1’, it must be in the last position. So the answer is “001”.
Example 2:

Input: s = “0101”
Output: “1001”
Explanation: One of the '1’s must be in the last position. The maximum number that can be made with the remaining digits is “100”. So the answer is “1001”.

Constraints:

1 <= s.length <= 100
s consists only of ‘0’ and ‘1’.
s contains at least one ‘1’.

二、题解

class Solution {
public:string maximumOddBinaryNumber(string s) {int n = s.size();int cnt = 0;for(int i = 0;i < n;i++){if(s[i] == '1') cnt++;}string res;for(int i = 0;i < cnt - 1;i++){res += "1";}for(int i = cnt - 1;i < n - 1;i++){res += "0";}res += "1";return res;}
};