动态规划——Best Time to Buy and Sell Stock III

题意：用一个数组表示股票每天的价格，数组的第i个数表示股票在第i天的价格。 如果最多进行两次交易，但必须在买进一只股票前清空手中的股票，求最大的收益。

示例 1:
Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
示例 2:
Input: [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.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.
示例 3:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

 

之前有一个Best Time to Buy and Sell Stock I ，比这个题简单点儿，那个题是最多一次操作，求最大的收益。同样是使用动态规划，由于这次允许最多两次操作，只需要分两次，

将输入的整个周期通过for对周期长度限制（第一个子周期长度递增且不超过原周期长）并分割成两个子周期先后进行dp数组的递推即可。

 

int maxProfit(int* prices, int pricesSize) {
     int ans = 0;
    if (pricesSize > 2){
        int*dp = (int*)malloc(pricesSize*sizeof(int));
        dp[0] = 0;
        int ans2 = 0;
        for (int i = 2; i <= pricesSize; i++){
            dp[0] = 0;
            ans2 = 0;
            for (int j = 1; j<i; j++){
                dp[j] = (prices[j] + dp[j - 1] - prices[j - 1]) > 0 ? (prices[j] + dp[j - 1] - prices[j - 1]) : 0;
                ans2 = ans2 >= dp[j] ? ans2 : dp[j];
            }
            ans = ans >= ans2 ? ans : ans2;
            if (i < pricesSize)dp[i] = 0;
            for (int k = i+1; k<pricesSize; k++){
                dp[k] = (prices[k] + dp[k - 1] - prices[k - 1]) > 0 ? (prices[k] + dp[k - 1] - prices[k - 1]) : 0;
                ans = ans >= (dp[k]+ans2) ? ans : (dp[k]+ans2);
            }
        }
        //free(dp);
    }
    else if (pricesSize == 2){
        ans = (prices[1] - prices[0]) > 0 ? (prices[1] - prices[0]):0;
    }
    else ans = 0;
    return ans;
}

但是非常搞笑的是我这个代码虽然在LeetCode上成功AC，但是应该是最差的一个解决方案。。。但是思路上非常便于理解