leetcode 539. Minimum Time Difference

Given a list of 24-hour clock time points in "Hour:Minutes" format, find the minimum minutes difference between any two time points in the list.

Example 1:

Input: ["23:59","00:00"]Output: 1

Note:

1. The number of time points in the given list is at least 2 and won't exceed 20000.
2. The input time is legal and ranges from 00:00 to 23:59.

我是想反正时间固定是时从0~23，分从0~59，那么我可以用一个数组来记录下。然后找这个数组中相邻两个之间差最小的。但是这没有考虑到数组中时间有重复的情况，所以在一开始存入数组中的时候，一旦发现重复的时间点，则返回0.

package leetcode;import java.util.ArrayList;import java.util.List;public class Minimum_Time_Difference_539 {public int findMinDifference(List<String> timePoints) {int[][] time=new int[24][60];int min=Integer.MAX_VALUE;for(String timeString:timePoints){String[] split=timeString.split(":");int hour=Integer.valueOf(split[0]);int minute=Integer.valueOf(split[1]);if(time[hour][minute]==1){return 0;}else{time[hour][minute]=1;}}int i,j=0;int[] before=new int[2],current=new int[2];boolean isFindBefore=false;int[] linchenfirst=new int[2];//保留凌晨第一个时间，用来与前一天最后一个时间相减for(i=0;i<24;i++){for(j=0;j<60;j++){if(time[i][j]==1){if(isFindBefore==false){before[0]=i;before[1]=j;linchenfirst[0]=i;linchenfirst[1]=j;isFindBefore=true;}else{current[0]=i;current[1]=j;int difference=difference(before, current);min=min<=difference?min:difference;before[0]=i;before[1]=j;}}}}int difference=difference(current, linchenfirst);min=min<=difference?min:difference;return min;}public int difference(int[] before,int[] current){if(before[0]==current[0]){return Math.abs(current[1]-before[1]);}else if(before[0]<current[0]){int minute=(current[0]-before[0])*60;if(before[1]<current[1]){minute+=current[1]-before[1];}else{minute-=before[1]-current[1];}return minute;}else{//一般current都会比before大的，这种情况应该是过了凌晨了current[0]+=24;return difference(before, current);}}public static void main(String[] args) {// TODO Auto-generated method stubMinimum_Time_Difference_539 m=new Minimum_Time_Difference_539();List<String> list=new ArrayList<String>();list.add("00:00");list.add("23:59");list.add("00:00");System.out.println(m.findMinDifference(list));}}

大神思路跟我一样，不过解法更妙更简洁。直接把时间算成hour*60+minute，就只有 24 * 60 = 1440 可能的时间点。那么可以用bucket sort，如果存在那个时间点，就在那个时间点的桶里放true.

public class Solution {    public int findMinDifference(List<String> timePoints) {        boolean[] mark = new boolean[24 * 60];        for (String time : timePoints) {            String[] t = time.split(":");            int h = Integer.parseInt(t[0]);            int m = Integer.parseInt(t[1]);            if (mark[h * 60 + m]) return 0;            mark[h * 60 + m] = true;        }                int prev = 0, min = Integer.MAX_VALUE;        int first = Integer.MAX_VALUE, last = Integer.MIN_VALUE;        for (int i = 0; i < 24 * 60; i++) {            if (mark[i]) {                if (first != Integer.MAX_VALUE) {                    min = Math.min(min, i - prev);                }                first = Math.min(first, i);                last = Math.max(last, i);                prev = i;            }        }                min = Math.min(min, (24 * 60 + first - last));                return min;    }}

first和last分别是记录数组中时间最小的时间点和时间最大的时间点，用来计算凌晨跨度的时间差。