力扣题:数字与字符串间转换-12.12

力扣题-12.12

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：539. 最小时间差

解题思想：将字符串的时间形式换成数字形式的时间，然后计算差值即可，最重要的是最小的值加上一天的时间加入到数组最后（计算第一个和最后一个时间的时间差）

class Solution(object):
    def findMinDifference(self, timePoints):
        """
        :type timePoints: List[str]
        :rtype: int
        """
        if len(timePoints)>24*60:
            return 0
        total = []
        for i in range(len(timePoints)):
            time = timePoints[i].split(":")
            minute = int(time[0])*60+int(time[1])
            total.append(minute)
        total = sorted(total)
        total.append(total[0]+24*60)
        result = 24*60
        for i in range(1,len(total)):
            result = min(result,total[i]-total[i-1])
        return result

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        if (timePoints.size() > 24 * 60) {
            return 0;
        }

        std::vector<int> total;
        for (const auto& timePoint : timePoints) {
            int colonIndex = timePoint.find(":");
            int hour = std::stoi(timePoint.substr(0, colonIndex));
            int minute = std::stoi(timePoint.substr(colonIndex + 1));
            int minutes = hour * 60 + minute;
            total.push_back(minutes);
        }

        std::sort(total.begin(), total.end());

        total.push_back(total[0] + 24 * 60);

        int result = 24 * 60;
        for (int i = 1; i < total.size(); ++i) {
            result = std::min(result, total[i] - total[i - 1]);
        }

        return result;
    }
};