refactor 1010

fishercoder1534 · fishercoder1534 · commit a8ec300f6aca · 2020-08-06T15:17:51.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_1010.java b/src/main/java/com/fishercoder/solutions/_1010.java
@@ -3,40 +3,15 @@
 import java.util.HashMap;
 import java.util.Map;
 
-/**
- * 1010. Pairs of Songs With Total Durations Divisible by 60
- *
- * In a list of songs, the i-th song has a duration of time[i] seconds.
- *
- * Return the number of pairs of songs for which their total duration in seconds is divisible by 60.
- * Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.
- *
- * Example 1:
- * Input: [30,20,150,100,40]
- * Output: 3
- * Explanation: Three pairs have a total duration divisible by 60:
- * (time[0] = 30, time[2] = 150): total duration 180
- * (time[1] = 20, time[3] = 100): total duration 120
- * (time[1] = 20, time[4] = 40): total duration 60
- *
- * Example 2:
- * Input: [60,60,60]
- * Output: 3
- * Explanation: All three pairs have a total duration of 120, which is divisible by 60.
- *
- * Note:
- *
- * 1 <= time.length <= 60000
- * 1 <= time[i] <= 500
- * */
 public class _1010 {
     public static class Solution1 {
-        /**Credit: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/discuss/256726/Java-O(n)-code-w-comment-similar-to-Two-Sum
-         *
+        /**
+         * Credit: https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/discuss/256726/Java-O(n)-code-w-comment-similar-to-Two-Sum
+         * <p>
          * Think of Problem 1: Two Sum
          * Assume target is 60, each item in time % 60.
          * Then this problem becomes very similar to Problem 1.
-         * */
+         */
         public int numPairsDivisibleBy60(int[] time) {
             int result = 0;
             Map<Integer, Integer> map = new HashMap<>();
