refactor 188

fishercoder1534 · fishercoder1534 · commit 43e30bb79987 · 2018-06-07T06:57:44.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_188.java b/src/main/java/com/fishercoder/solutions/_188.java
@@ -1,46 +1,58 @@
 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.
+
+ 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 {
-
-    /**credit: https://discuss.leetcode.com/topic/8984/a-concise-dp-solution-in-java*/
+  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]);
-            }
+      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];
-            }
+      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;
     }
-
+  }
 }
