《LeetCode之每日一题》:255.一周中的第几天

题目链接： 一周中的第几天

有关题目

给你一个日期，请你设计一个算法来判断它是对应一周中的哪一天。

输入为三个整数：day、month 和 year，分别表示日、月、年。

您返回的结果必须是这几个值中的一个 {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}。

示例 1：

输入：day = 31, month = 8, year = 2019
输出："Saturday"

示例 2：

输入：day = 18, month = 7, year = 1999
输出："Sunday"

示例 3：

输入：day = 15, month = 8, year = 1993
输出："Sunday"

提示：

给出的日期一定是在 1971 到 2100 年之间的有效日期。

题解

法一：模拟
参考官方题解
代码一：

class Solution {
public:
    string dayOfTheWeek(int day, int month, int year) {
        vector<string> weekdays = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};
        vector<int> months = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int days = 0;

        //统计输入年份之前的天数贡献
        for (int i = 1971; i < year; i++)
        {
            if ((i % 400 == 0) ||( i % 4 == 0 && i % 100 != 0))
                days += 366;
            else 
                days += 365;
        }

        bool check = (year % 400 == 0) ||( year % 4 == 0 && year % 100 != 0);
        if (check)//year为闰年则二月多加一天
        {
            months[1]++;
        }
        for (int i = 0; i < month - 1; i++)
        {
            days += months[i];
        }

        days += day;
        //1971.1.1当天为周五
        return weekdays[(days + 3) % 7];
    }
};

代码二：

class Solution {
public:
    string dayOfTheWeek(int day, int month, int year) {
        vector<string> weekdays = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};
        vector<int> months = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        //统计输入年份之前的天数贡献
        int days = 365 * (year - 1971) + (year - 1969) / 4;//四年一闰   

        for (int i = 0; i < month - 1; i++)
        {
            days += months[i];
        }

        //year为闰年且月份超过2月则，总天数加 1
        if ((year % 400 == 0 ||( year % 4 == 0 && year % 100 != 0)) && month > 2)
        {
            days += 1;
        }
        
        days += day;
        //1971.1.1当天为周五
        return weekdays[(days + 3) % 7];
    }
};