> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/l7shu-zu/best-time-to-buy-and-sell-stock-iv.md).

# Best Time to Buy and Sell Stock IV

Say you have an array for which theithelement is the price of a given stock on dayi.

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).

分析

也是维护global max和local max

如果k>=n/2以后，就贪心法做

```
class Solution {
    public int maxProfit(int k, int[] prices) {        
        int n = prices.length;       
        if(n< 1)
            return 0;

        if (k >=  n/2) {
            int maxPro = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i-1])
                    maxPro += prices[i] - prices[i-1];
            }
            return maxPro;
        }

        int localMax = 0;
        int[][] f = new int[n + 1][k + 1];
        for(int j = 1; j <= k; j ++){
             localMax = f[0][j - 1] - prices[0];//max需要初始化！！！
            for(int i = 1; i <= n; i ++){
            //这次卖或者不卖，卖的话所以加
                f[i][j] = Math.max(f[i - 1][j], prices[i - 1] + localMax);
            //之前某次买的，所以要减，这次保存了Local就能省一次loop，有loop的话这里就是f[k][j - 1] - prices[k - 1]k次买的
                localMax = Math.max(f[i][j - 1] - prices[i - 1], localMax);
            }
        }

        return f[n][k];
        }

}
```