Sword finger offer II 035 Minimum time difference

Summary

Use a length of 60*24=1440 Array of , The point in time when the marker appears . Then calculate the distance between the two .

subject

Given a 24 hourly （ Hours : minute “HH:MM”） Time list for , Find the minimum time difference between any two times in the list and express it in minutes .

link ：https://leetcode.cn/problems/569nqc/

Ideas

Time list is 24 hourly , And accurate to points .24 Hours , Yes 1440 minute .

We can use a length of 1440 Array of , As a hash table . Drop every minute into every grid . Then traverse each lattice , Record the shortest distance between grids with values . Of course , Don't forget the difference between the earliest and the latest time of the day , Also participate in the calculation .

solution ： length 1440 Array of records time

Code

public int findMinDifference(List<String> timePoints) {
    int[] list = new int[1440];
    for (String timePoint : timePoints) {
        //  Have the same time point , Go straight back to 0
        int hash = hash(timePoint);
        if (list[hash] == 1) {
            return 0;
        }
        list[hash]++;
    }
    int min = Integer.MAX_VALUE;
    int start = 0;
    int first = 0;
    //  Initialize start node 
    for (int i = 0; i < list.length; i++) {
        if (list[i] != 0) {
            start = i;
            first = i;
            break;
        }
    }

    for (int i = start + 1; i < list.length; i++) {
        if (list[i] > 0) {
            min = Math.min(min, i - start);
            start = i;
        }
    }

    return Math.min(min, list.length - start + first);
}

private int hash(String timePoint) {
    String[] split = timePoint.split(":");
    return Integer.parseInt(split[0]) * 60 + Integer.parseInt(split[1]);

}