From 6bc2b03cf80c5ba859deb06709e0c33a9366dfbd Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Sun, 2 Sep 2018 17:44:57 +0800
Subject: [PATCH] Create 122. Best Time to Buy and Sell Stock II.md

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+# [122. Best Time to Buy and Sell Stock II](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/)
+# 思路
+## 比较好想的思路
+如果分别知道了前0到前i天的最大利润dp[0...i]，那么前i+1天的最大利润就为：  
+max(dp[j] + max(prices[i+1] - prices[k])), 其中k属于j~i+1   
+以上思路时间复杂度为O(n^2), 空间复杂度为O(n)  
+## 改进
+运用贪心的思想：只要有利润(即prices[i] > prices[i-1])就可以买入卖出, 即不错过任何利润。  
+此时时间复杂度O(n), 空间复杂度为O(1)
+
+# C++
+改进前：
+```
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        if(prices.size() == 0) return 0;
+        int dp[prices.size()] = {0};
+        int min_price;
+        for(int i = 1; i < prices.size(); i++){
+            min_price = prices[i];
+            for(int j = i-1; j >= 0; j--){
+                if(min_price > prices[j]) min_price = prices[j];
+                if(dp[j] + prices[i] - min_price > dp[i]) dp[i] = dp[j] + prices[i] - min_price;    
+            }
+        }
+        return dp[prices.size() - 1];
+    }
+};
+```
+改进后：
+```
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int max_profit = 0;
+        for(int i = 1; i < prices.size(); i++)
+            if(prices[i] > prices[i-1]) max_profit += (prices[i] - prices[i-1]);
+        return max_profit;
+    }
+};
+```