Leetcode Best Time to Buy and Sell Stock 2-白红宇

Leetcode Best Time to Buy and Sell Stock 2

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发布时间：2019-05-26

本文共 1543 字，大约阅读时间需要 5 分钟。

/************************************************************************

* 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(ie, buy one and sell one share of the stock multiple times).However,

*you may not engage in multiple transactions at the same time(ie, you must sell the

* stock before you buy again).

**********************************************************************/

题目解析：例如股票的变化如下

23,24,27,18,23,19,31

这里我们可以随意操作股票，因此，只要将股票下降的前一天，立即收回股票，

明天继续入手。

因此可分在第一天入手，第三天出手

第四天入手，第五天出手

第六天入手，第七天出手

// Author : yqtao// Date   : 2016-6.19// Email  ：yqtao@whu.edu.cn#include "stdafx.h"#include 
   
    #include 
    
     using namespace std;int maxProfit(vector
     
       &price){    int begin = 0, end = 0, max = 0;    //这里添加了两个标志，判断股票状态    bool up = false, down = false;    for (int i = 1; i
      
        price[i - 1] && up == false) { // goes up            begin = i - 1;            up = true;            down = false;        }        if (price[i] < price[i - 1] && down == false) { // goes down            end = i - 1;            down = true;            up = false;            max += (price[end] - price[begin]);        }    }    // 考虑边界条件    if (begin < price.size() && up == true) {        end = price.size() - 1;        max += (price[end] - price[begin]);    }        return max;}int main(){    vector
       
        price = { 23,24,27,18,23,19,31 };    cout << maxProfit(price) << endl;}