refactor 188

fishercoder1534 · fishercoder1534 · commit 7aa3ac6bc932 · 2020-10-18T07:24:52.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_188.java b/src/main/java/com/fishercoder/solutions/_188.java
@@ -1,58 +1,37 @@
 package com.fishercoder.solutions;
 
-/**
-
- 188. Best Time to Buy and Sell Stock IV
-
- Say you have an array for which the ith element is the price of a given stock on day i.
-
- Design an algorithm to find the maximum profit. You may complete at most k transactions.
-
- Note:
- You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
-
- Example 1:
- Input: [2,4,1], k = 2
- Output: 2
- Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
-
- Example 2:
- Input: [3,2,6,5,0,3], k = 2
- Output: 7
- Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
- Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
-
- */
 public class _188 {
-  public static class Solution1 {
-    /** credit: https://discuss.leetcode.com/topic/8984/a-concise-dp-solution-in-java */
-    public int maxProfit(int k, int[] prices) {
-      int len = prices.length;
-      if (k >= len / 2) {
-        return quickSolve(prices);
-      }
-
-      int[][] t = new int[k + 1][len];
-      for (int i = 1; i <= k; i++) {
-        int tmpMax = -prices[0];
-        for (int j = 1; j < len; j++) {
-          t[i][j] = Math.max(t[i][j - 1], prices[j] + tmpMax);
-          tmpMax = Math.max(tmpMax, t[i - 1][j - 1] - prices[j]);
+    public static class Solution1 {
+        /**
+         * credit: https://discuss.leetcode.com/topic/8984/a-concise-dp-solution-in-java
+         */
+        public int maxProfit(int k, int[] prices) {
+            int len = prices.length;
+            if (k >= len / 2) {
+                return quickSolve(prices);
+            }
+
+            int[][] t = new int[k + 1][len];
+            for (int i = 1; i <= k; i++) {
+                int tmpMax = -prices[0];
+                for (int j = 1; j < len; j++) {
+                    t[i][j] = Math.max(t[i][j - 1], prices[j] + tmpMax);
+                    tmpMax = Math.max(tmpMax, t[i - 1][j - 1] - prices[j]);
+                }
+            }
+            return t[k][len - 1];
         }
-      }
-      return t[k][len - 1];
-    }
 
-    private int quickSolve(int[] prices) {
-      int len = prices.length;
-      int profit = 0;
-      for (int i = 1; i < len; i++) {
-        // as long as there is a price gap, we gain a profit.
-        if (prices[i] > prices[i - 1]) {
-          profit += prices[i] - prices[i - 1];
+        private int quickSolve(int[] prices) {
+            int len = prices.length;
+            int profit = 0;
+            for (int i = 1; i < len; i++) {
+                // as long as there is a price gap, we gain a profit.
+                if (prices[i] > prices[i - 1]) {
+                    profit += prices[i] - prices[i - 1];
+                }
+            }
+            return profit;
         }
-      }
-      return profit;
     }
-  }
 }
