[leetcode]539. Minimum Time Difference
阿新 • • 發佈:2018-11-09
[leetcode]539. Minimum Time Difference
Analysis
今天要吃柚子—— [每天刷題並不難0.0]
Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.
先對輸入的字串陣列排序,然後遍歷一下算差值,最後再取最小的差值就可以了。
Implement
class Solution {
public:
int findMinDifference(vector<string>& timePoints) {
int res = INT_MAX;
int len = timePoints.size();
sort(timePoints.begin(), timePoints.end());
int h1, h2, m1, m2;
int diff;
string time1, time2;
for(int i=0; i<len; i+ +){
time1 = timePoints[i];
time2 = timePoints[(i+1)%len];
h1 = (time1[0]-'0')*10+(time1[1]-'0');
m1 = (time1[3]-'0')*10+(time1[4]-'0');
h2 = (time2[0]-'0')*10+(time2[1]-'0');
m2 = (time2[3]-'0')*10+(time2[4]-'0');
diff = (h2-h1)*60 +(m2-m1);
if(i == len-1)
diff += 24*60;
res = min(res, diff);
}
return res;
}
};