# 188.Best-Time-to-Buy-and-Sell-Stock-IV

## 188. Best Time to Buy and Sell Stock IV

## 题目地址

<https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/>

<https://www.lintcode.com/problem/best-time-to-buy-and-sell-stock-iv/description>

## 题目描述

```
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).

Example 1:
Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:
Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
```

## 代码

### Approach #1 Dynamic Programming

```java
public class Solution {
  public int maxProfit(int k, int[] prices) {
    if (prices == null || prices.length <= 1 || k <= 0) return 0;
        // 1. Special case
    int n = prices.length;
    if (k >= n / 2) {
      int profit_max = 0;
      for (int i = 1; i < n; i++) {
        if (prices[i] - prices[i - 1] > 0) {
          profit_max += prices[i] - prices[i - 1];
        }
      }
      return profit_max;
    }
        // 2. dp
    int[][] f = new int[n + 1][k + 1];
    for (int j = 1; j <= k; j++) {
      for (int i = 1; i <= n; i++) {
        for (int x = 0; x <= i; x++) {
          f[i][j] = Math.max(f[i][j], f[x][j - 1] + profit(prices, x + 1, i));
        }
      }
    }
    return f[n][k];
  }

  // [left, right]一次交易最大收益
  private int profit(int[] prices, int left, int right) {
    if (left >= right) return 0;

    int leftMin = Integer.MAX_VALUE;
    int profit = 0;
    for (int i = left; i < right; i++) {
      profit = Math.max(profit, prices[i] - leftMin);
      leftMin = Math.min(leftMin, prices[i]);
    }
    return profit;
  }
}
```


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