refactor 1353

fishercoder1534 · fishercoder1534 · commit 56b196e1e0d7 · 2020-10-05T06:44:01.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_1353.java b/src/main/java/com/fishercoder/solutions/_1353.java
@@ -1,59 +1,18 @@
 package com.fishercoder.solutions;
 
-import com.fishercoder.common.utils.CommonUtils;
-
 import java.util.Arrays;
-import java.util.Collections;
 import java.util.PriorityQueue;
 
-/**
- * 1353. Maximum Number of Events That Can Be Attended
- *
- * Given an array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayi and ends at endDayi.
- * You can attend an event i at any day d where startTimei <= d <= endTimei. Notice that you can only attend one event at any time d.
- * Return the maximum number of events you can attend.
- *
- * Example 1:
- * Input: events = [[1,2],[2,3],[3,4]]
- * Output: 3
- * Explanation: You can attend all the three events.
- * One way to attend them all is as shown.
- * Attend the first event on day 1.
- * Attend the second event on day 2.
- * Attend the third event on day 3.
- *
- * Example 2:
- * Input: events= [[1,2],[2,3],[3,4],[1,2]]
- * Output: 4
- *
- * Example 3:
- * Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
- * Output: 4
- *
- * Example 4:
- * Input: events = [[1,100000]]
- * Output: 1
- *
- * Example 5:
- * Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
- * Output: 7
- *
- * Constraints:
- * 1 <= events.length <= 10^5
- * events[i].length == 2
- * 1 <= events[i][0] <= events[i][1] <= 10^5
- * */
 public class _1353 {
     public static class Solution1 {
         /**
          * Credit: https://leetcode.com/problems/maximum-number-of-events-that-can-be-attended/discuss/510263/JavaC%2B%2BPython-Priority-Queue
-         *
+         * <p>
          * 1. Sort events by start time, if ties, by end time;
          * 2. From day 1 to day 100,000, we add all events that start on this day into a priorityqueue,
          * also, we remove the events that closed on this day from the priorityqueue;
          * 3. attend the event that ends on this day (earliest, i.e. greedy) and pop it out of the priorityqueue
-         *
-         * */
+         */
         public int maxEvents(int[][] events) {
             Arrays.sort(events, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
             PriorityQueue<Integer> heap = new PriorityQueue<>();
