Best Time to Buy and Sell Stock II (E)
題目
Say you have an array prices for which the ith 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 (i.e., buy one and sell one share of the stock multiple times).
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: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 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.
Example 3:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Constraints:
1 <= prices.length <= 3 * 10 ^ 40 <= prices[i] <= 10 ^ 4
題意
在一個時刻買入股票，在另一個時刻賣出股票，可以多次交易，求能得到的最大收益，
思路
動態規劃：和 0309. Best Time to Buy and Sell Stock with Cooldown (M) 方法一樣，
One Pass：從左到右遍歷陣列，只要后一天的價格比前一天高，那么就可以把這份利潤加入到總利潤中，但這并不是說可以賣出后又在同一天買入，例如：[1, 2, 3]，按照我們的方法看起來像是 1買入->2賣出->2買入->3賣出，實際上的交易只有 1買入->3賣出，中間的部分只是數學意義上的變通，
代碼實作
Java
動態規劃
class Solution {
public int maxProfit(int[] prices) {
int[] sold = new int[prices.length];
int[] hold = new int[prices.length];
hold[0] = -prices[0];
for (int i = 1; i < prices.length; i++) {
hold[i] = Math.max(hold[i - 1], sold[i - 1] - prices[i]);
sold[i] = Math.max(sold[i - 1], hold[i - 1] + prices[i]);
}
return sold[prices.length - 1];
}
}
One Pass
class Solution {
public int maxProfit(int[] prices) {
int sum = 0;
for (int i = 0; i < prices.length - 1; i++) {
sum += prices[i + 1] > prices[i] ? prices[i + 1] - prices[i] : 0;
}
return sum;
}
}
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