minor edits

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_396.java

@@ -1,6 +1,9 @@
 package com.fishercoder.solutions;
 
-/**Given an array of integers A and let n to be its length.
+/**
+ * 396. Rotate Function
+ *
+ * Given an array of integers A and let n to be its length.
 
  Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:
 
@@ -21,6 +24,7 @@ Calculate the maximum value of F(0), F(1), ..., F(n-1).
  F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
 
  So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.*/
+
 //F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
 public class _396 {
     public int maxRotateFunction(int[] A) {
@@ -54,7 +58,7 @@ private int[] rotate(int[] a) {
     }
 
     //**credit : https://discuss.leetcode.com/topic/58459/java-o-n-solution-with-explanation
-    public int maxRotateFunction_1(int[] A) {
+    public int maxRotateFunctionV2(int[] A) {
         int allSum = 0;
         int len = A.length;
         int F = 0;
@@ -74,6 +78,6 @@ public static void main(String...strings){
         int[] nums = new int[]{4, 3, 2, 6};
         _396 test = new _396();
         System.out.println(test.maxRotateFunction(nums));
-        System.out.println(test.maxRotateFunction_1(nums));
+        System.out.println(test.maxRotateFunctionV2(nums));
     }
 }

src/main/java/com/fishercoder/solutions/_42.java

@@ -1,38 +1,39 @@
 package com.fishercoder.solutions;
 
 /**
+ * 42. Trapping Rain Water
  * Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
 
  For example,
  Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
 
-
- The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
+ The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
  */
+
 public class _42 {
 
     public int trap(int[] height) {
         int len = height.length;
-        if(len == 0){
+        if (len == 0) {
             return 0;
         }
         int res = 0;
 
-        //first use DP to calculate out left and right arrays
+        //first use DP to calculate left and right arrays
         int[] left = new int[height.length];
         int[] right = new int[height.length];
 
         left[0] = height[0];
-        for(int i = 1; i < height.length; i++){
-            left[i] = Math.max(left[i-1], height[i]);
+        for (int i = 1; i < height.length; i++) {
+            left[i] = Math.max(left[i - 1], height[i]);
         }
 
-        right[height.length-1] = height[height.length-1];
-        for(int i = height.length-2; i >= 0; i--){
-            right[i] = Math.max(right[i+1], height[i]);
+        right[height.length - 1] = height[height.length - 1];
+        for (int i = height.length - 2; i >= 0; i--) {
+            right[i] = Math.max(right[i + 1], height[i]);
         }
 
-        for(int i = 1; i < height.length-1; i++){
+        for (int i = 1; i < height.length - 1; i++) {
             res += Math.min(left[i], right[i]) - height[i];
         }
         return res;