Question
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 k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
Solution
This question is very difficult. We need to do DP with 2 DP arrays, available to read here.
The 2 arrays' definition as follow:
global[i][j]=max(local[i][j],global[i-1][j])
当前到达第i天可以最多进行j次交易，最好的利润是多少（global[i][j]）
当前到达第i天，最多可进行j次交易，并且最后一次交易在当天卖出的最好的利润是多少（local[i][j]）
And the formula for calculating local[] is:
local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff)，
第一个是全局到i-1天进行j-1次交易，然后加上今天的交易，如果今天是赚钱的话（也就是前面只要j-1次交易，最后一次交易取当前天），
第二个量则是取local第i-1天j次交易，然后加上今天的差值。
And the final code (by blogger Code_Ganker from the same link) would look like this:
public int maxProfit(int[] prices) {
if(prices==null || prices.length==0)
return 0;
int[] local = new int[3];
int[] global = new int[3];
for(int i=0;i<prices.length-1;i++)
{
int diff = prices[i+1]-prices[i];
for(int j=2;j>=1;j--)
{
local[j] = Math.max(global[j-1]+(diff>0?diff:0), local[j]+diff);
global[j] = Math.max(local[j],global[j]);
}
}
return global[2];
}