[LeetCode]309. Best Time to Buy

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 as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

• You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
• After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

##### 方法
sell[i]表示第i天卖掉股票的最大收益,`sell[i]>=sell[i-1]`
buy[i]表示第i天买了股票后的最大收益
有如下公式：
```c
buy[i] = max(buy[i-1], sell[i-2]-prices[i]); 
sell[i] = max(buy[i-1]+prices[i], sell[i-1]);

c代码

#include <assert.h>
#include <stdlib.h>

#define max(A,B) ((A)>(B)?(A):(B))

int maxProfit(int* prices, int pricesSize) {
    int i = 0;
    int* buy = (int *)malloc(sizeof(int) * pricesSize);
    int* sell = (int *)malloc(sizeof(int) * pricesSize);
    buy[0] = -prices[0];
    buy[1] = prices[1]>prices[0]?-prices[0]:-prices[1];
    sell[0] = 0;
    sell[1] = prices[1]-prices[0];
    sell[1] = sell[1]>0?sell[1]:0;

    for(i = 2; i < pricesSize; i++) {
        buy[i] = max(buy[i-1], sell[i-2]-prices[i]);
        sell[i] = max(buy[i-1]+prices[i], sell[i-1]);
    }
    return sell[i-1];
}

int main() {
    int nums[5] = {1,2,3,0,2};
    assert(maxProfit(nums, 5) == 3);

    return 0;
}