https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock. Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Example 1: Input: [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
- code
class Solution {
public int maxProfit(int[] prices) {
int res = 0, last = prices[0], cur = 0;
for (int i = 1; i < prices.length; i++){
cur += prices[i] - prices[i-1];
if (cur < 0){
cur = 0;
continue;
}
res = Math.max(cur, res);
}
return res;
}
}
- code
public class Solution {
public int maxProfit(int prices[]) {
int minprice = Integer.MAX_VALUE;
int maxprofit = 0;
for (int i = 0; i < prices.length; i++) {
if (prices[i] < minprice)
minprice = prices[i];
else if (prices[i] - minprice > maxprofit)
maxprofit = prices[i] - minprice;
}
return maxprofit;
}
}
- code
class Solution:
def maxProfit(self, prices: List[int]) -> int:
cur = res = 0
for i in range(1, len(prices)):
cur += prices[i] - prices[i-1]
if cur < 0: cur = 0
res = max(res, cur)
return res
- code
class Solution:
def maxProfit(self, prices: List[int]) -> int:
minday = prices[0]
maxprofit = 0
for v in prices:
if v < minday:
minday = v
else:
maxprofit = max(maxprofit, v - minday)
return maxprofit