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.
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。
1 int maxProfit(int* prices, int pricesSize) { 2 int ans = 0; 3 if (pricesSize > 2){ 4 int*dp = (int*)malloc(pricesSize*sizeof(int)); 5 dp[0] = 0; 6 int ans2 = 0; 7 for (int i = 2; i <= pricesSize; i++){ 8 dp[0] = 0; 9 ans2 = 0; 10 for (int j = 1; j<i; j++){ 11 dp[j] = (prices[j] + dp[j - 1] - prices[j - 1]) > 0 ? (prices[j] + dp[j - 1] - prices[j - 1]) : 0; 12 ans2 = ans2 >= dp[j] ? ans2 : dp[j]; 13 } 14 ans = ans >= ans2 ? ans : ans2; 15 if (i < pricesSize)dp[i] = 0; 16 for (int k = i+1; k<pricesSize; k++){ 17 dp[k] = (prices[k] + dp[k - 1] - prices[k - 1]) > 0 ? (prices[k] + dp[k - 1] - prices[k - 1]) : 0; 18 ans = ans >= (dp[k]+ans2) ? ans : (dp[k]+ans2); 19 } 20 } 21 //free(dp); 22 } 23 else if (pricesSize == 2){ 24 ans = (prices[1] - prices[0]) > 0 ? (prices[1] - prices[0]):0; 25 } 26 else ans = 0; 27 return ans; 28 }
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
