《LeetCode之每日一题》:270.最小时间差

Posted 2022-02-07 是七喜呀！

最小时间差

• 有关题目
• • 题解

题目链接： 最小时间差

有关题目

给定一个 24 小时制（小时:分钟 "HH:MM"）的时间列表，
找出列表中任意两个时间的最小时间差并以分钟数表示。

示例 1：

输入：timePoints = ["23:59","00:00"]
输出：1

示例 2：

输入：timePoints = ["00:00","23:59","00:00"]
输出：0

提示：

2 <= timePoints.length <= 2 * 10^4
timePoints[i] 格式为 "HH:MM"

题解

法一：排序
参考官方题解

#define MIN(a, b) ((a) < (b) ? (a) : (b)) 


int getMinutes(const char *s)

    int a = s[0] - '0', b = s[1] - '0';
    int h  = a * 10 + b;
    int c = s[3] - '0', d = s[4] - '0';
    int m = c * 10 + d;
    return h * 60 + m;


int cmp(const void *e1, const void *e2)//升序排序

    return strcmp(*(char **)e1, *(char **)e2);


int findMinDifference(char ** timePoints, int timePointsSize)
    qsort(timePoints, timePointsSize, sizeof(char*), cmp);
    int t0Minutes = getMinutes(timePoints[0]);
    int preMinutes = t0Minutes;
    int ans = 1440;//24小时最多的分钟数
    for (int i = 1; i < timePointsSize; i++)
    
        int minutes = getMinutes(timePoints[i]);
        ans = MIN(ans, minutes - preMinutes);//相邻时间差
        preMinutes = minutes;
    

    ans = MIN(ans, t0Minutes + 1440 - preMinutes);//首位时间差
    
    return ans;

法二：鸽巢原理
参考官方题解

#define MIN(a, b) ((a) < (b) ? (a) : (b)) 


int getMinutes(const char *s)

    int a = s[0] - '0', b = s[1] - '0';
    int h  = a * 10 + b;
    int c = s[3] - '0', d = s[4] - '0';
    int m = c * 10 + d;
    return h * 60 + m;


int cmp(const void *e1, const void *e2)//升序排序

    return strcmp(*(char **)e1, *(char **)e2);


int findMinDifference(char ** timePoints, int timePointsSize)
    //鸽巢原理,总共右24 * 60 = 1440中时间，超过这个数量一定会有重复的时间,最小时间差为0
    if(timePointsSize > 1440)
        return 0;

    qsort(timePoints, timePointsSize, sizeof(char*), cmp);
    int t0Minutes = getMinutes(timePoints[0]);
    int preMinutes = t0Minutes;
    int ans = 1440;//24小时最多的分钟数
    for (int i = 1; i < timePointsSize; i++)
    
        int minutes = getMinutes(timePoints[i]);
        ans = MIN(ans, minutes - preMinutes);//相邻时间差
        preMinutes = minutes;
    

    ans = MIN(ans, t0Minutes + 1440 - preMinutes);//首位时间差
    
    return ans;