Leetcode Best Time to Buy and Sell Stock III 123

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).

题目链接

要求最多操作两次，而且只能在上次交易完成之后，才可以进行第二次交易

考虑从左，从右分别进行遍历，找到对于当前的最大利润

另外，有可能不操作或者操作一次，所以还需要遍历一下只进行一次的交易

class Solution {
public:
int maxProfit(vector<int>& prices) {
int len=prices.size();
if(len==0 || len==1) return 0;
if(len==2){
if(prices[1]>prices[0]) return prices[1]-prices[0];
else return 0;
}
vector<int> lsum(len+1,0);
vector<int> rsum(len+1,0);

int min_=prices[0];
int mp=0;
int maxprice=0;

for(int i=0;i<len;i++){
if(prices[i]<min_) min_=prices[i];
mp=prices[i]-min_;
if(mp>maxprice) maxprice=mp;
lsum[i]=maxprice;
}

int max_=prices[len-1];
mp=0;
maxprice=0;
for(int i=len-1;i>=0;i--){
if(prices[i]>max_) max_=prices[i];
mp=max_-prices[i];
if(mp>maxprice) maxprice=mp;
rsum[i]=maxprice;
}

int temp=lsum[0]+rsum[1];
for(int i=1;i<len-1;i++){
if(temp<(lsum[i]+rsum[i+1])){
temp=lsum[i]+rsum[i+1];
}
}

for(int i=0;i<len;i++){
if(temp<lsum[i]) temp=lsum[i];
if(temp<rsum[i]) temp=rsum[i];
}
return temp;

}
};