标签：epo begin nim 取值 += find vector time rip
给定一个 24 小时制（小时:分钟）的时间列表，找出列表中任意两个时间的最小时间差并已分钟数表示。
示例 1：
输入: ["23:59","00:00"]
输出: 1
备注:
1.列表中时间数在 2~20000 之间。
2.每个时间取值在 00:00~23:59 之间。
详见：https://leetcode.com/problems/minimum-time-difference/description/
C++：
class Solution {
public:
int findMinDifference(vector<string>& timePoints)
{
int res = INT_MAX, n = timePoints.size(), diff = 0;
sort(timePoints.begin(), timePoints.end());
for (int i = 0; i < n; ++i)
{
string t1 = timePoints[i], t2 = timePoints[(i + 1) % n];
int h1 = (t1[0] - ‘0‘) * 10 + t1[1] - ‘0‘;
int m1 = (t1[3] - ‘0‘) * 10 + t1[4] - ‘0‘;
int h2 = (t2[0] - ‘0‘) * 10 + t2[1] - ‘0‘;
int m2 = (t2[3] - ‘0‘) * 10 + t2[4] - ‘0‘;
diff = (h2 - h1) * 60 + (m2 - m1);
if (i == n - 1)
{
diff += 24 * 60;
}
res = min(res, diff);
}
return res;
}
};
参考：http://www.cnblogs.com/grandyang/p/6568398.html
539 Minimum Time Difference 最小时间差
标签：epo begin nim 取值 += find vector time rip
原文地址：https://www.cnblogs.com/xidian2014/p/8910267.html
