123. Best Time to Buy and Sell Stock III
You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete at most two transactions.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [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: prices = [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.
Example 3:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Example 4:
Input: prices = [1]
Output: 0
Constraints:
- 1 <= prices.length <= 105
- 0 <= prices[i] <= 105
class Solution {
public int maxProfit(int[] prices) {
if(prices.length==0){
return 0;
}
int b1 = Integer.MAX_VALUE;
int b2 = Integer.MAX_VALUE;
int p1=0, p2=0;
for(int i=0;i<prices.length;i++){
// 最低价格 = 找出最低的价格
b1 = Math.min(b1,prices[i]);
// 最大收益 = 当前价格-最低的价格
p1 = Math.max(p1,prices[i]-b1);
// 当前价格-以往最高的收益
b2 = Math.min(b2,prices[i]-p1);
// 当前价格 + 以往最高收益
p2 = Math.max(p2,prices[i]-b2);
}
return p2;
}
}