refactor 330

fishercoder1534 · fishercoder1534 · commit 8c9eff85ff87 · 2020-06-04T07:48:57.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_330.java b/src/main/java/com/fishercoder/solutions/_330.java
@@ -3,42 +3,17 @@
 import java.util.ArrayList;
 import java.util.List;
 
-/**
- * 330. Patching Array
- *
- * Given a sorted positive integer array nums and an integer n,
- * add/patch elements to the array such that any number in range [1, n]
- * inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.
-
- Example 1:
- nums = [1, 3], n = 6
- Return 1.
-
- Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4.
- Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3].
- Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6].
- So we only need 1 patch.
-
- Example 2:
- nums = [1, 5, 10], n = 20
- Return 2.
- The two patches can be [2, 4].
-
- Example 3:
- nums = [1, 2, 2], n = 5
- Return 0.
- */
 public class _330 {
 
     public static class Solution1 {
         /**
          * credit: https://leetcode.com/articles/patching-array/ and https://discuss.leetcode.com/topic/35494/solution-explanation/2
-         *
+         * <p>
          * Let miss be the smallest sum in [0,n] that we might be missing. Meaning we already know we
          * can build all sums in [0,miss). Then if we have a number num <= miss in the given array, we
          * can add it to those smaller sums to build all sums in [0,miss+num). If we don't, then we must
          * add such a number to the array, and it's best to add miss itself, to maximize the reach.
-         *
+         * <p>
          * Example: Let's say the input is nums = [1, 2, 4, 13, 43] and n = 100. We need to ensure that
          * all sums in the range [1,100] are possible. Using the given numbers 1, 2 and 4, we can
          * already build all sums from 0 to 7, i.e., the range [0,8). But we can't build the sum 8, and
