?

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?two?transactions.（股票交易，最多兩次）

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

?

class Solution { public:int maxProfit(vector<int> &prices) {if(prices.size()==0)return 0;int n=prices.size();vector<int> left(n);vector<int> right(n);int min=prices[0];int max=prices[n-1];int res=0;for(int i=1;i<n;i++){min=min<prices[i]?min:prices[i];left[i]=left[i-1]>(prices[i]-min)?left[i-1]:(prices[i]-min);}for(int j=n-2;j>=0;j--){max=max>prices[j]?max:prices[j];right[j]=right[j+1]>(max-prices[j])?right[j+1]:(max-prices[j]);}for(int i=0;i<n;i++){res=res>(left[i]+right[i])?res:(left[i]+right[i]);}return res;} };

　　

轉(zhuǎn)載于:https://www.cnblogs.com/Vae1990Silence/p/4830586.html

與50位技術(shù)專家面對(duì)面20年技術(shù)見證，附贈(zèng)技術(shù)全景圖

總結(jié)

以上是生活随笔為你收集整理的最多两次股票交易-Best Time to Buy and Sell Stock III的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。