LeetCode 解題筆記：2409. Count Days Spent Together

作者：王一哲
日期：2026年10月24日


LeetCode 題目連結：2409. Count Days Spent Together

解題想法

簡單題。題目給 4 個長度皆為 5 的字串，格式為 MM-DD，MM 代表月，DD 代表日。4 個字串分別是代表 Alice 抵達的日期 arriveAlice，Alice 離開的日期 leaveAlice，Bob 抵達的日期 arriveBob，Bob 離開的日期 leaveBob。計算 Alice 與 Bob 一起待在羅馬的天數。假設是平年，2 月只有 28 天。

由於日期格式很困定，可以用 Python 的字串切片或是 C++ 的 substr，將字串中的月、日分開，轉成整數。如果用 C 語言比較麻煩，我是用索引值取出字元再減去 '0' 換成對應的整數，每個位數都要分開算。在計算天數時，我將日期全部轉成由1月1日開始的累計天數。為了計算月份對應的天數，我開一個長度為 13 的陣列 $month = (0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365)$，記錄經過完整的月份時共經過幾天。答案為兩個人離開日期最小值 - 兩個人離開日期最大值，如果答案是負的則回傳 0。

Python 程式碼

Runtime: 0 ms, beats 100.00%. Memory: 19.29 MB, beats 79.30%.

class Solution:
    def countDaysTogether(self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str) -> int:
        month = (0, 31, 59, 90, 120, 151, 181,
                 212, 243, 273, 304, 334, 365)
        arriveAliceDays = month[int(arriveAlice[:2]) - 1] + int(arriveAlice[3:])
        leaveAliceDays = month[int(leaveAlice[:2]) - 1] + int(leaveAlice[3:])
        arriveBobDays = month[int(arriveBob[:2]) - 1] + int(arriveBob[3:])
        leaveBobDays = month[int(leaveBob[:2]) - 1] + int(leaveBob[3:])
        ans = min(leaveAliceDays, leaveBobDays) - max(arriveAliceDays, arriveBobDays) + 1
        return ans if ans > 0 else 0

C++ 程式碼

Runtime: 0 ms, beats 100.00%. Memory: 8.05 MB, beats 38.18%.

class Solution {
public:
    int countDaysTogether(string arriveAlice, string leaveAlice, string arriveBob, string leaveBob) {
        int month[13] = 
            {0, 31, 59, 90, 120, 151, 181,
             212, 243, 273, 304, 334, 365};
        int arriveAliceDays = month[stoi(arriveAlice.substr(0, 2)) - 1] + stoi(arriveAlice.substr(3));
        int leaveAliceDays = month[stoi(leaveAlice.substr(0, 2)) - 1] + stoi(leaveAlice.substr(3));
        int arriveBobDays = month[stoi(arriveBob.substr(0, 2)) - 1] + stoi(arriveBob.substr(3));
        int leaveBobDays = month[stoi(leaveBob.substr(0, 2)) - 1] + stoi(leaveBob.substr(3));
        int ans = min(leaveAliceDays, leaveBobDays) - max(arriveAliceDays, arriveBobDays) + 1;
        return (ans > 0 ? ans : 0);
    }
};

C 語言程式碼

Runtime: 0 ms, beats 100.00%. Memory: 8.48 MB, beats 62.50%.

#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))

int countDaysTogether(char* arriveAlice, char* leaveAlice, char* arriveBob, char* leaveBob) {
    int month[13] = 
        {0, 31, 59, 90, 120, 151, 181,
         212, 243, 273, 304, 334, 365};
    int arriveAliceDays = month[(arriveAlice[0] - '0') * 10 + (arriveAlice[1] - '0') - 1] + (arriveAlice[3] - '0') * 10 + (arriveAlice[4] - '0');
    int leaveAliceDays = month[(leaveAlice[0] - '0') * 10 + (leaveAlice[1] - '0') - 1] + (leaveAlice[3] - '0') * 10 + (leaveAlice[4] - '0');
    int arriveBobDays = month[(arriveBob[0] - '0') * 10 + (arriveBob[1] - '0') - 1] + (arriveBob[3] - '0') * 10 + (arriveBob[4] - '0');
    int leaveBobDays = month[(leaveBob[0] - '0') * 10 + (leaveBob[1] - '0') - 1] + (leaveBob[3] - '0') * 10 + (leaveBob[4] - '0');
    int ans = min(leaveAliceDays, leaveBobDays) - max(arriveAliceDays, arriveBobDays) + 1;
    return (ans > 0 ? ans : 0);
}