1154.一年中的第几天

链接：1154.一年中的第几天
难度：Easy
标签：数学、字符串
简介：给你一个字符串 date ，按 YYYY-MM-DD 格式表示一个 现行公元纪年法 日期。请你计算并返回该日期是当年的第几天。

题解 1 - cpp

• 编辑时间：2021-12-21
• 执行用时：16ms
• 内存消耗：5.8MB
• 编程语言：cpp
• 解法介绍：检测闰年和前面有几个月。

class Solution {
   public:
    int monthDay[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    void checkLeapYear(int year) {
        if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {
            monthDay[2] = 29;
        }
    }
    int dayOfYear(string date) {
        int year = 0, month = 0, day = 0;
        for (int i = 0; i < 4; i++) year = year * 10 + date[i] - '0';
        for (int i = 5; i < 7; i++) month = month * 10 + date[i] - '0';
        for (int i = 8; i < 10; i++) day = day * 10 + date[i] - '0';
        checkLeapYear(year);
        // printf("year = %d, month = %d, day = %d", year, month, day);
        int ans = day;
        for (int i = 1; i < month; i++) ans += monthDay[i];
        return ans;
    }
};

题解 2 - python

• 编辑时间：2023-12-31
• 执行用时：48ms
• 内存消耗：17.11MB
• 编程语言：python
• 解法介绍：直接计算。

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 dayOfYear(self, date: str) -> int:
                year, month, day = int(date[:4]), int(date[5:7]), int(date[8:])
                months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30]
                if isLeapYear(year): months[1] = 29
                return sum(months[:month - 1]) + day