# 539.Minimum-Time-Difference

## 539. Minimum Time Difference

## 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

```java
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

```java
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
}
```


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