LeetCode188. 买卖股票的最佳时机 IV

188. 买卖股票的最佳时机 IV

思路

四种状态 当天买入, 不买, 卖出, 不卖，状态转移方程

buy[i] = max(buy[i], sell[i-1] - prices)    //buy[i]代表第i笔买入自己还剩的钱 买入则减去当天的价格 

sell[i] = max(sell[i], buy[i] + prices[i])   //sell[i]代表第i笔卖出后自己还剩的钱 卖出即加入当天的价格

参考代码

class Solution {
    public int maxProfit(int k, int[] prices) {
        if (prices == null || prices.length == 0) return 0;
        if (k >= prices.length / 2){
            int res = 0;
            for (int i = 1; i < prices.length; i++)
                res += Math.max(0, prices[i] - prices[i-1]);
            return res;
        }
        int []buy = new int[k+1];
        int []sell  = new int[k+1];
        Arrays.fill(buy, Integer.MIN_VALUE);
        for(int i = 0;i < prices.length; i++){   
            for (int j = 0; j < k; j++) {
                buy[j+1] = Math.max(buy[j+1], sell[j] - prices[i]);
                sell[j+1] = Math.max(buy[j+1] + prices[i], sell[j+1]);
            }
        }
        return sell[k];
    }
}