Given an array of 4 digits, return the largest 24 hour time that can be made.

The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight.

Return the answer as a string of length 5. If no valid time can be made, return an empty string.

Example 1:

Input: [1,2,3,4]
Output: "23:41"

Example 2:

Input: [5,5,5,5]
Output: ""

Note:

A.length == 4
0 <= A[i] <= 9

Solution

Time complexity : O(1)
Space complexity : O(1)

class Solution {
public:
    string largestTimeFromDigits(vector<int>& A) {
        int total_min = INT_MIN;
        
        sort(A.begin(), A.end());
        do {
            total_min = max(total_min, cal_total_min(A));
        } while (next_permutation(A.begin(), A.end()));
        
        string res = "";
        if (total_min != INT_MIN) {
            string str_min = to_string(total_min%60);
            if (str_min.length() < 2) str_min = "0" + str_min;
            string str_hour = to_string(total_min/60);
            if (str_hour.length() < 2) str_hour = "0" + str_hour;
            res = str_hour + ":" + str_min;
        }
        
        return res;
    }
private:
    bool is_valid_time(vector<int> vec) {
        if (vec[0]>2 ||
            vec[0]==2 && vec[1]>3 ||
            vec[2]>5) {
            return false;
        } else return true;
    }
    
    int cal_total_min(vector<int> vec) {
        if (!is_valid_time(vec)) return INT_MIN;
        return (vec[0]*10+vec[1])*60+(vec[2]*10+vec[3]);
    }
};

思路很簡單，所有排列組合、合法的時間中，取最大者。
有些麻煩的是要做時間合法性判斷、計算總時間量、還有字串處理。