1185.一周中的第几天

链接：1185.一周中的第几天
难度：Easy
标签：数学
简介：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。

题解 1 - python

• 编辑时间：2023-12-30
• 执行用时：36ms
• 内存消耗：17.09MB
• 编程语言：python
• 解法介绍：计算天数后取模。

def isLeapYear(year: int) -> bool:
        return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0
    weeks = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
    
    class Solution:
        def dayOfTheWeek(self, day: int, month: int, year: int) -> str:
            months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30]
            if isLeapYear(year): months[1] = 29
            day += sum(365 + isLeapYear(y) for y in range(1971, year)) + sum(months[:month - 1])
            return weeks[(day + 3) % 7]

题解 2 - cpp

• 编辑时间：2022-01-04
• 内存消耗：5.9MB
• 编程语言：cpp
• 解法介绍：遍历求出距离第一天的天数并取模。

string names[] = {"Friday",  "Saturday",  "Sunday",  "Monday",
                  "Tuesday", "Wednesday", "Thursday"};
int isLeapYear(int year) {
    return year % 400 == 0 || year % 100 != 0 && year % 4 == 0;
}
class Solution {
   public:
    string dayOfTheWeek(int day, int month, int year) {
        int months[] = {
            0,  31, isLeapYear(year) ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31,
            30, 31};
        while (month) day += months[--month];
        while (year > 1971) day += isLeapYear(--year) ? 366 : 365;
        return names[(day - 1) % 7];
    }
};