539.Minimum-Time-Difference.md

539. Minimum Time Difference

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题目地址

https://leetcode.com/problems/minimum-time-difference/

题目描述

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

代码

Approach #1 Sort

class Solution {
  public int findMinDifference(List<String> timePoints) {
		int mm = Integer.MAX_VALUE;
    List<Integer> time = new ArrayList<>();
    
    for (int i = 0; i < timePoints.size(); i++) {
      Integer h = Integer.valueOf(timePoints.get(i).substring(0, 2));
      time.add(60 * h + Integer.valueOf(timePoints.get(i).substring(3, 5)));
    }
    
    Collections.sort(time, (Integer a, Integer b) -> a - b);
    
    for (int i = 1; i < time.size(); i++) {
      mm = Math.min(mm, time.get(i) - time.get(i - 1));
    }
    
    int corner = time.get(0) + (1440 - time.get(time.size() - 1));
    return Math.min(corner, mm);
  }
}

Approach #2 Priority

public int findMinDifference(List<String> timePoints) {
	PriorityQueue<Integer> pq = new PriorityQueue<>();
	for (String s : timePoints) {
		int h = Integer.valueOf(s.substring(0,2));
		int m = Integer.valueOf(s.substring(3));
		pq.offer(h*60+m);
	}
	if (pq.size() < 2) return 0;
	int res = Integer.MAX_VALUE, first = pq.poll();
	int cur = first;
	while (!pq.isEmpty()) {
		int next = pq.poll();
		res = Math.min(res, next-cur);
		cur = next;
	}
	return Math.min(res, 24*60 - cur + first); // difference
}