README.md

1154. 一年中的第几天

English Version

题目描述

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。返回该日期是当年的第几天。

 

示例 1：

输入：date = "2019-01-09"
输出：9
解释：给定日期是2019年的第九天。

示例 2：

输入：date = "2019-02-10"
输出：41

 

提示：

• date.length == 10
• date[4] == date[7] == '-'，其他的 date[i] 都是数字
• date 表示的范围从 1900 年 1 月 1 日至 2019 年 12 月 31 日

解法

闰年 2 月有 29 天，平年 2 月有 28 天。

闰年的判断规则：year % 100 == 0 || (year % 4 == 0 && year % 100 != 0)

Python3

class Solution:
    def dayOfYear(self, date: str) -> int:
        year, month, day = (int(e) for e in date.split('-'))
        d = 29 if year % 400 == 0 or (year % 4 == 0 and year % 100 != 0) else 28
        days = [31, d, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
        return sum(days[: month - 1]) + day

Java

class Solution {
    public int dayOfYear(String date) {
        int year = Integer.parseInt(date.substring(0, 4));
        int month = Integer.parseInt(date.substring(5, 7));
        int day = Integer.parseInt(date.substring(8));
        int d = year % 400 == 0 || (year % 4 == 0 && year % 100 != 0) ? 29 : 28;
        int[] days = new int[] {31, d, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int ans = day;
        for (int i = 0; i < month - 1; ++i) {
            ans += days[i];
        }
        return ans;
    }
}

C++

class Solution {
public:
    int dayOfYear(string date) {
        int year = stoi(date.substr(0, 4));
        int month = stoi(date.substr(5, 7));
        int day = stoi(date.substr(8));
        int d = year % 400 == 0 || (year % 4 == 0 && year % 100 != 0) ? 29 : 28;
        int days[] = {31, d, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int ans = day;
        for (int i = 0; i < month - 1; ++i) ans += days[i];
        return ans;
    }
};

Go

func dayOfYear(date string) int {
	year, _ := strconv.Atoi(date[:4])
	month, _ := strconv.Atoi(date[5:7])
	day, _ := strconv.Atoi(date[8:])
	days := []int{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
	if year%400 == 0 || (year%4 == 0 && year%100 != 0) {
		days[1]++
	}
	ans := day
	for i := 0; i < month-1; i++ {
		ans += days[i]
	}
	return ans
}

JavaScript

/**
 * @param {string} date
 * @return {number}
 */
var dayOfYear = function (date) {
    const year = +date.slice(0, 4);
    const month = +date.slice(5, 7);
    const day = +date.slice(8);
    const d =
        year % 400 === 0 || (year % 4 === 0 && year % 100 !== 0) ? 29 : 28;
    const days = [31, d, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
    let ans = day;
    for (let i = 0; i < month - 1; ++i) {
        ans += days[i];
    }
    return ans;
};

...