Best Time to Buy and Sell Stock IV

可以买卖k次，DP来解决，

t[i][j] = Math.max(t[i][j - 1], prices[j] + buy) 即得出在prices[j]卖出时的最大利润; 

buy= Math.max(buy, t[i - 1][j - 1] - prices[j]) 则得出在prices[j]买进后剩下的钱，留待下次卖出.

 

当k大于等于len/2时，问题转化为可以买卖任意次的Best Time to Buy and Sell Stock II ，不然会很不efficient，会超时。 

当k=2时，则转化为买卖两次的Best Time to Buy and Sell Stock III 。 

 

参考链接： 链接一 

而链接二 中的local，global变量的方法与 Maximum Subarray 一样，但是不好理解。

 

public class Solution {
    public int maxProfit(int k, int[] prices) {
        int len = prices.length;
        if (k > len / 2) return quickSolve(prices);
        int[][] dp = new int[k + 1][len];
        for(int i = 1; i <= k; i++) {
            int buy = -prices[0];
            for(int j = 1; j < len; j++) {
                dp[i][j] = Math.max(dp[i][j - 1], prices[j] + buy);
                buy = Math.max(buy, dp[i - 1][j - 1] - prices[j]);
            }
        }
        return dp[k][len - 1];
    }


    private int quickSolve(int[] prices) {
        int len = prices.length, profit = 0;
        for (int i = 1; i < len; i++)
            // as long as there is a price gap, we gain a profit.
            if (prices[i] > prices[i - 1]) profit += prices[i] - prices[i - 1];
        return profit;
    }
}