[LeetCode10]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).

Analysis

One dimensional dynamic planning.the max profit at day i is the max profit before day i + max profit after day i.
Given an i, split the whole array into two parts:
[0,i] and [i+1, n], it generates two max value based on i, Max(0,i) and Max(i+1,n)
So, we can define the transformation function as:
Maxprofix = max(Max(0,i) + Max(i+1, n))  0<=i<n

Java

public int maxProfit(int[] prices) {        int maxPro = 0;if(prices.length>1){ArrayList<Integer> mp = new ArrayList<>();int minP = prices[0];mp.add(maxPro);for(int i=1;i<prices.length;i++){if(prices[i]-minP>=maxPro){ maxPro = prices[i]-minP;}if(minP>prices[i]) minP = prices[i];mp.add(maxPro);}maxPro =0;int maxP = prices[prices.length-1];for(int i=prices.length-2;i>=0;i--){if(maxPro<maxP-prices[i]+mp.get(i)){maxPro = maxP-prices[i]+mp.get(i);}if(prices[i]>maxP) maxP = prices[i];}}return maxPro;    }

c++

int maxProfit(vector<int> &prices){    if(prices.size()<=1) return 0;    if(prices.size()==2) return prices[1]>prices[0]?prices[1]-prices[0]:0;    vector<int> maxFromLeft(prices.size(),0);    vector<int> maxFromRight(prices.size(),0);    int minL = INT_MAX, max=INT_MIN, diff;    for(int i=0;i<prices.size();i++){        if(prices[i]<minL) minL = prices[i];        diff = prices[i]- minL;        if(diff >max)            max = diff;        maxFromLeft[i] = max;    }    int maxR = INT_MIN;    max = INT_MIN;    for(int i=prices.size()-1;i>=0;i--){        if(prices[i]>maxR) maxR = prices[i];        diff = maxR - prices[i];        if(diff > max)            max = diff;        maxFromRight[i] = max;    }    max = INT_MIN;    for(int i=0;i<prices.size()-1;i++){        int diff = maxFromLeft[i]+maxFromRight[i+1];        if(diff > max) max = diff;    }    if(max < maxFromRight[0])        max = maxFromRight[0];    return max;}