文章目录
1185. 一周中的第几天
取模
利用一种模运算的性质:对长度(在本题就是一周的天数也就是7)取模得到的余数就是在一个周期内相差的(例如本题就是用总相差的天数对7取模得到的数字就是要求的这一天在一个星期内和第一天相差几天)。
于是,转换成了一道求天数的题
1.考虑到题中所给年份从1971开始,所以我把第一天设为1971.1.1(星期五)
2.先求整年中有多少天,先把所有年都当作平年计算,然后遍历发现有闰年天数+1
3.求整月的天数,先用一个数组存下来每个月的天数,然后循环加就可以了(这里注意判断一下这一年是不是闰年,2月的天数是多少)
4.后面day的值直接加就好啦
class Solution {
public:
string dayOfTheWeek(int day, int month, int year) {
int q[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
string s[]={"Friday","Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday"};
int cnt=(year-1971)*365;
for(int i=1971;i<year;i++)
{
if((i%4==0&&i%100!=0)||i%400==0) cnt++;
}
if((year%4==0&&year%100!=0)||year%400==0) q[2]++;
for(int i=1;i<month;i++) cnt+=q[i];
cnt+=day-1;
int qi=cnt%7;
return s[qi];
}
};
1154. 一年中的第几天
暴力求解
我这一题的做法和上一题求天数的做法相同,甚至核心代码都是复制来的,稍微有点麻烦的是输入。温馨提示:力扣不能用atoi,只能用stoi
class Solution {
public:
int dayOfYear(string date) {
int cnt=0;
int year = stoi(date.substr(0, 4));
int month = stoi(date.substr(5, 2));
int day = stoi(date.substr(8, 2));
int q[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
if((year%4==0&&year%100!=0)||year%400==0) q[2]++;
for(int i=1;i<month;i++) cnt+=q[i];
cnt+=day;
return cnt;
}
};
1360. 日期之间隔几天
[题目链接](https://leetcode-cn.com/problems/number-of-days-between-two-dates/)
cv编辑法,复制第一题代码
class Solution {
public:
int dayOfTheWeek(int day, int month, int year) {
int q[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int cnt=(year-1971)*365;
for(int i=1971;i<year;i++)
{
if((i%4==0&&i%100!=0)||i%400==0) cnt++;
}
if((year%4==0&&year%100!=0)||year%400==0) q[2]++;
for(int i=1;i<month;i++) cnt+=q[i];
cnt+=day-1;
return cnt;
}
int daysBetweenDates(string date1, string date2) {
int year1 = stoi(date1.substr(0, 4));
int month1 = stoi(date1.substr(5, 2));
int day1 = stoi(date1.substr(8, 2));
int year2 = stoi(date2.substr(0, 4));
int month2 = stoi(date2.substr(5, 2));
int day2 = stoi(date2.substr(8, 2));
int ans1=dayOfTheWeek(day1,month1,year1);
int ans2=dayOfTheWeek(day2,month2,year2);
return abs(ans2-ans1);
}
};