[LeetCode] Best Time to Buy and Sell Stock III

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).

本题我用的方法同样比较麻烦，欢迎大家进行交流~

第一步：合并相同的点

第二步：找到全部的极小值和极大值，先找极小值再找极大值

第三步：第二步找到 mx 组极值，将这 mx 组分割成两部分，分别找这两个部分的最大利润，想当于在两个部分内分别计算Best Time to Buy and Sell Stock I的问题

class Solution {
public:
int maxProfit(vector<int> &prices) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int size=prices.size();
if(size==0) return 0;
int *newPrices=new int[size];
int *maxPrices=new int[size];
int *minPrices=new int[size];
int n=0;
int mx=0;
int mn=0;
for(int s=0;s<size;s++){//合并相同的点
if(s<size-1&&prices[s]==prices[s+1]) continue;
newPrices
=prices[s];
n++;
}
if(newPrices[0]<newPrices[1]){// 分析首位置的情况
minPrices[mn]=newPrices[0];
mn++;
}
for(int i=1;i<n-1;i++){
if(newPrices[i]>newPrices[i-1]&&newPrices[i]>newPrices[i+1]){
maxPrices[mx]=newPrices[i];
mx++;
}
if(newPrices[i]<newPrices[i-1]&&newPrices[i]<newPrices[i+1]){
minPrices[mn]=newPrices[i];
mn++;
}
}
if(newPrices[n-1]>newPrices[n-2]){//分析末尾的情况
maxPrices[mx]=newPrices[n-1];
mx++;
}
else{
minPrices[mn]=newPrices[n-1];
mn++;
}
int sum_min=0;
int sum_max=0;
if(mx==0) return 0;
if(mx==1) return (maxPrices[0]-minPrices[0]);
if(mx==2) return (maxPrices[1]+maxPrices[0]-minPrices[1]-minPrices[0]);
int max=0;

for(int m=1;m<mx;m++)
{
int temp_max1=maxPrices[0]-minPrices[0];
int minPos1=0;
for(int i=0;i<m;i++)
{
if(minPrices[i]<minPrices[minPos1]) minPos1=i;
if(maxPrices[i]-minPrices[minPos1]>temp_max1)
temp_max1=maxPrices[i]-minPrices[minPos1];

}
int temp_max2=maxPrices[m]-minPrices[m];
int minPos2=m;
for(int j=m;j<mx;j++)
{
if(minPrices[j]<minPrices[minPos2]) minPos2=j;
if(maxPrices[j]-minPrices[minPos2]>temp_max2)
temp_max2=maxPrices[j]-minPrices[minPos2];

}
if(temp_max2+temp_max1>max) max=temp_max2+temp_max1;

//cout<<m<<" "<<temp_max1<<" "<<temp_max2<<endl;
}

delete []newPrices;
delete []maxPrices;
delete []minPrices;
return max;
}
};