# 122. Best Time to Buy and Sell Stock II (M)

You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day.

On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can buy it then immediately sell it on the **same day**.

Find and return *the **maximum** profit you can achieve*.

 

**Example 1:**

```
Input: prices = [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.
```

**Example 2:**

```
Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.
```

**Example 3:**

```
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
```

 

**Constraints:**

* `1 <= prices.length <= 3 * 104`
* `0 <= prices[i] <= 104`

### Solution:

如果 `k` 为正无穷，那么就可以认为 `k` 和 `k - 1` 是一样的。可以这样改写框架：

```python
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i])
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0] - prices[i])
            = max(dp[i-1][k][1], dp[i-1][k][0] - prices[i])

我们发现数组中的 k 已经不会改变了，也就是说不需要记录 k 这个状态了：
dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i])
dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i])
```

直接翻译成代码：

```java
// 原始版本
int maxProfit_k_inf(int[] prices) {
    int n = prices.length;
    int[][] dp = new int[n][2];
    for (int i = 0; i < n; i++) {
        if (i - 1 == -1) {
            // base case
            dp[i][0] = 0;
            dp[i][1] = -prices[i];
            continue;
        }
        dp[i][0] = Math.max(dp[i-1][0], dp[i-1][1] + prices[i]);
        dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] - prices[i]);
    }
    return dp[n - 1][0];
}

// 空间复杂度优化版本
int maxProfit_k_inf(int[] prices) {
    int n = prices.length;
    int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++) {
        int temp = dp_i_0;
        dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
        dp_i_1 = Math.max(dp_i_1, temp - prices[i]);
    }
    return dp_i_0;
}
```


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