Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.
You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid.
Example 1:
Input: "19:34" Output: "19:39" Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later. It is not 19:33, because this occurs 23 hours and 59 minutes later.
Example 2:
Input: "23:59" Output: "22:22" Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22. It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.
A:
就是逐个增加,然后看是否存在呗
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | class Solution { public: string nextClosestTime(string time) { vector<bool> V(10,false); for(int i =0;i<5;i++){ if(i==2) continue; V[ time[i]-'0'] = true; } int hh = stoi(time.substr(0,2)); int mm = stoi(time.substr(3,2)); for(int len = 1; len<=60*24; len++){ int m = (mm + len % 60); // might have carry int h = (hh + len / 60 + m/60) %24 ; m = m%60; if(V[h/10] && V[h%10] && V[m/10] && V[m%10]){ return to_string(h/10) + to_string(h%10) + ":" + to_string(m/10) + to_string(m%10); } } return ""; } }; |
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