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;
}