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）

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class Solution(object):def findMinDifference(self, timePoints):""":type timePoints: List[str]:rtype: int"""if len(timePoints)>24*60:return 0total = []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*60for 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;}
};