[LeetCode] Best Time to Buy and Sell Stock II

标签：

This problem is similar to the one Best time to sell and buy stock. To maximize the profit, in each time interval [a, b], we choose day i such that prices[i] is the local minimum ofprices[a..b] and day j such that prices[j] is the local maximum of prices[a..b]. Then the max profit in interval [a, b] is prices[j] - prices[i] if j > i or simply 0 otherwise.

Suppose we are given n days [1..n], we need to find the first local minimum (supposed to bebuy) and the first local maximum (called sell), then if buy < sell, the transaction is valid and we earn the profit, after which we update both buy and sell to be the starting point of the next interval sell + 1. Otherwise if buy > sell, we need to update sell to be at least buy + 1(updating sell to buy is meaningless since it will give profit of 0, which can be handled by initializing max_profit to be 0).

Well, now comes the following code, with comments.

 1     int maxProfit(vector<int>& prices) {
 2         int max_profit = 0, buy = 0, sell = 0;
 3         while (buy < prices.size() && sell < prices.size()) {
 4             if (buy < prices.size() - 1 && prices[buy] > prices[buy + 1]) buy++; // move buy to local minimum
 5             else if (sell < prices.size() - 1 && prices[sell] < prices[sell + 1]) sell++; // move sell to local maximum
 6             else 
 7                 if (buy < sell){ 
 8                     // the transaction is valid, take it and update buy and sell to be the starting point of the next interval
 9                     max_profit += prices[sell] - prices[buy];
10                     buy = ++sell;
11                 }
12                 else sell = buy + 1; // the date to sell need to be after the day to buy
13         }
14         return max_profit;
15     }

[LeetCode] Best Time to Buy and Sell Stock II

标签：

原文地址：http://www.cnblogs.com/jcliBlogger/p/4548108.html