[LeetCode 188] Best Time to Buy and Sell Stock IV

Question

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Say you have an array for which the i^th 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.

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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];
}