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.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
给定一个数组prices, prices[i]表示第i天的股价。本题规定最多只能买卖两次，问最大收益是多少
class Solution {
public:
int maxProfit(vector<int> &prices) {
int size=prices.size();
if(size<=1)return 0;
int*back=new int[size];
int*front=new int[size];
int maxProfit=0;
int minPrice=prices[0];
int maxPrice=prices[size-1];
back[size-1]=front[0]=0;
// 求出以i结尾的前半段区间上买卖一次可获得最大收益
maxProfit=0;
for(int i=1; i<size; i++){
int profit=prices[i]-minPrice;
if(profit>maxProfit)maxProfit=profit;
front[i]=maxProfit;
if(prices[i]<minPrice)minPrice=prices[i];
}
// 求出以i开始的后半段区间上买卖一次可获得最大收益
maxProfit=0;
for(int i=size-2; i>=0; i--){
int profit=maxPrice-prices[i];
if(profit>maxProfit)maxProfit=profit;
back[i]= maxProfit;
if(prices[i]>maxPrice)maxPrice=prices[i];
}
//求两次买卖的最大值
maxProfit=0;
for(int i=0; i<size; i++){
if(i==size-1){
if(front[i]>maxProfit)maxProfit=front[i];
}
else{
if(front[i]+back[i+1]>maxProfit)maxProfit=front[i]+back[i+1];
}
}
return maxProfit;
}
};LeetCode: Best Time to Buy and Sell Stock III [123],布布扣,bubuko.com
LeetCode: Best Time to Buy and Sell Stock III [123]
原文：http://blog.csdn.net/harryhuang1990/article/details/33345117