买股票题目一种很好的通用解法

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/?tab=Description

题目是最多两次，但是下面的解法非常好，也能够覆盖仅仅一次的情况。

https://discuss.leetcode.com/topic/5934/is-it-best-solution-with-o-n-o-1/2

代码很简短，思路也非常好：

public class Solution {
    public int maxProfit(int[] prices) {
        int hold1 = Integer.MIN_VALUE, hold2 = Integer.MIN_VALUE;
        int release1 = 0, release2 = 0;
        for(int i:prices){                              // Assume we only have 0 money at first
            release2 = Math.max(release2, hold2+i);     // The maximum if we've just sold 2nd stock so far.
            hold2    = Math.max(hold2,    release1-i);  // The maximum if we've just buy  2nd stock so far.
            release1 = Math.max(release1, hold1+i);     // The maximum if we've just sold 1nd stock so far.
            hold1    = Math.max(hold1,    -i);          // The maximum if we've just buy  1st stock so far. 
        }
        return release2; ///Since release1 is initiated as 0, so release2 will always higher than release1.
    }
}

但是上面对于最多N次这种更加通用的形式，就不太好用了。

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/?tab=Description

上面对于最多买n次类型的题目，要用到DP才行。

解法看这里：

https://discuss.leetcode.com/topic/8984/a-concise-dp-solution-in-java

其实含义也很明确：

tmpMax是当前时间状态持股之后的最大收益（很可能是负的）

t[i][j]是记录的实际收益。

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