123. Best Time to Buy and Sell Stock III
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 at most two transactions.
**Note: **You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
Example 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.
Example 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.
思路：
这一题也是买卖股票的问题，条件是允许买卖两次，最直接的办法是在121的基础上，分为两次最大盈利做。这里只说一下动态规划来解。
代码：
java：
代码语言：javascript复制class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) return 0;
int[] dp = new int[3];
int[] min = new int[]{prices[0], prices[0], prices[0]};
for (int i = 0; i < prices.length; i ) {
for (int k = 1; k <= 2; k ) {
dp[k] = Math.max(dp[k], prices[i] - min[k]);
min[k] = Math.min(min[k], prices[i] - dp[k-1]);
}
}
return dp[2];
}
}
