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539.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
}