获得最小时差 [英] Getting minimum time difference

问题描述

最近在一次编程采访中有人问我这个问题.我用HH:MM格式给出了一系列的时间.我需要找到以分钟为单位的最小时间差.时间本质上是循环的，因此23:55和00:12应当相差17.该功能还需要像24:00一样处理00:00(自然这是最简单的部分).有人可以帮助我开始解决这个问题.

I was asked this question recently in a coding interview. I'm given a series of times in HH:MM format. I need to find the minimum time difference in minutes. The time is circular in nature, thus 23:55 and 00:12 should have a difference of 17. The function also needs to handle 00:00 as the same thing as 24:00(this is the easiest part, naturally). Can someone help me get started with the problem.

推荐答案

您可以尝试以下方法:

import java.util.Arrays;
public class Main {
    public static void main(String args[]) {
        String times[] = {"2:30","00:00", "00:12", "24:00", "23:55"};/*input times*/
        int countTime = times.length;
        int sortedTimes[] = new int[countTime];/*will hold sorted times*/

        for (int i = 0; i < countTime; i++) {/*convert 'HH:mm' to integer minute*/
            int t = Integer.valueOf(times[i].split(":")[0])*60 
                    + Integer.valueOf(times[i].split(":")[1]);
            sortedTimes[i] = t;
        }
        Arrays.sort(sortedTimes);/*sort times*/         
        /*variables to hold the result*/
        int minDiff = 1440;
        int minT1 = 1440;
        int minT2 = 1440;

        for (int i = 0; i < countTime; i++) {
            int curr = sortedTimes[i];
            int prev = sortedTimes[(countTime+i-1)%countTime];
            int diffPrev = Math.abs(curr - prev) < 1440 - Math.abs(curr - prev ) ? 
                    Math.abs(curr - prev) :  1440 - Math.abs(curr - prev  ) ; 
            if (minDiff > diffPrev) {
                    minDiff = diffPrev;
                    minT1 = prev;
                    minT2 = curr;                   
            }
        }       
        System.out.println("Minimum diff in minute: " + minDiff 
                + " between " + (minT1/60 + ":" + minT1%60) 
                + " and " + (minT2/60 + ":" + minT2%60));

    }
}

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