finish "122. Best Time to Buy and Sell Stock II". Fix #122

Author: diguage

Diff for: README.adoc

@@ -614,11 +614,11 @@
 |{source_base_url}/BestTimeToBuyAndSellStock.java[Java]
 |Easy
 
-//|122
-//|{leetcode_base_url}/best-time-to-buy-and-sell-stock-ii/[Best Time to Buy and Sell Stock II]
-//|{source_base_url}/BestTimeToBuyAndSellStockIi.java[Java]
-//|Easy
-//
+|122
+|{leetcode_base_url}/best-time-to-buy-and-sell-stock-ii/[Best Time to Buy and Sell Stock II]
+|{source_base_url}/BestTimeToBuyAndSellStockIi.java[Java]
+|Easy
+
 //|123
 //|{leetcode_base_url}/best-time-to-buy-and-sell-stock-iii/[Best Time to Buy and Sell Stock III]
 //|{source_base_url}/BestTimeToBuyAndSellStockIii.java[Java]

Diff for: src/main/java/com/diguage/algorithm/leetcode/BestTimeToBuyAndSellStockIi.java

@@ -0,0 +1,110 @@
+package com.diguage.algorithm.leetcode;
+
+import java.util.Objects;
+
+/**
+ * = 122. Best Time to Buy and Sell Stock II
+ *
+ * https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/[Best Time to Buy and Sell Stock II - LeetCode]
+ *
+ * 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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
+ *
+ * Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
+ *
+ * .Example 1:
+ * [source]
+ * ----
+ * Input: [7,1,5,3,6,4]
+ * Output: 7
+ * Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
+ *              Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
+ * ----
+ *
+ * .Example 2:
+ * [source]
+ * ----
+ * Input: [1,2,3,4,5]
+ * Output: 4
+ * Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
+ *              Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
+ *              engaging multiple transactions at the same time. You must sell before buying again.
+ * ----
+ *
+ * .Example 3:
+ * [source]
+ * ----
+ * Input: [7,6,4,3,1]
+ * Output: 0
+ * Explanation: In this case, no transaction is done, i.e. max profit = 0.
+ * ----
+ *
+ * @author D瓜哥, https://www.diguage.com/
+ * @since 2020-01-04 10:55
+ */
+public class BestTimeToBuyAndSellStockIi {
+
+    /**
+     * Runtime: 1 ms, faster than 85.09% of Java online submissions for Best Time to Buy and Sell Stock II.
+     *
+     * Memory Usage: 37 MB, less than 100.00% of Java online submissions for Best Time to Buy and Sell Stock II.
+     */
+    public int maxProfit(int[] prices) {
+        if (Objects.isNull(prices) || prices.length == 0) {
+            return 0;
+        }
+        int result = 0;
+        for (int i = 1; i < prices.length; i++) {
+            if (prices[i - 1] < prices[i]) {
+                result += prices[i] - prices[i - 1];
+            }
+        }
+        return result;
+    }
+    /**
+     * Runtime: 1 ms, faster than 85.09% of Java online submissions for Best Time to Buy and Sell Stock II.
+     *
+     * Memory Usage: 37 MB, less than 100.00% of Java online submissions for Best Time to Buy and Sell Stock II.
+     *
+     * 我的思路，别人的代码。
+     */
+    public int maxProfitPeakValleyApproach(int[] prices) {
+        if (Objects.isNull(prices) || prices.length == 0) {
+            return 0;
+        }
+        int i = 0;
+        int valley = prices[0];
+        int peak = prices[0];
+        int result = 0;
+        while (i < prices.length - 1) {
+            while (i < prices.length - 1 && prices[i] > prices[i + 1]) {
+                i++;
+            }
+            valley = prices[i];
+            while (i < prices.length - 1 && prices[i] < prices[i++]) {
+                i++;
+            }
+            peak = prices[i];
+            result += peak - valley;
+        }
+
+        return result;
+    }
+
+    public static void main(String[] args) {
+        BestTimeToBuyAndSellStockIi solution = new BestTimeToBuyAndSellStockIi();
+
+        int[] a1 = {7, 1, 5, 3, 6, 4};
+        int r1 = solution.maxProfit(a1);
+        System.out.println((r1 == 7) + " : " + r1);
+
+        int[] a2 = {1, 2, 3, 4, 5};
+        int r2 = solution.maxProfit(a2);
+        System.out.println((r2 == 4) + " : " + r2);
+
+        int[] a3 = {7, 6, 4, 3, 1};
+        int r3 = solution.maxProfit(a3);
+        System.out.println((r3 == 0) + " : " + r3);
+    }
+}