Code & Fun

第49天。

今天的题目是Best Time to Buy and Sell Stock:

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

之前好像做过一道类似的题目,但是那道题比这道题难多了,那道题是可以多次买入卖出的,而成每次交易是需要支付一定费用的,这道就简单多了,我们只需要记录当前最小元素,然后每次更新最小元素,然后记录当前元素与最小元素的差值即可。

1
2
3
4
5
6
7
8
9
10
11
12
13
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
profit,minElem = 0,sys.maxsize
for p in prices:
if p < minElem:
minElem = p
t = p - minElem
if t > profit:
profit = t
return profit

然后是c++的解:

1
2
3
4
5
6
7
8
9
int maxProfit(vector<int> &prices) {
int maxPro = 0;
int minPrice = INT_MAX;
for(int i = 0; i < prices.size(); i++){
minPrice = min(minPrice, prices[i]);
maxPro = max(maxPro, prices[i] - minPrice);
}
return maxPro;
}

以及在dicuss中看到的:

1
2
3
4
5
6
7
8
public int maxProfit(int[] prices) {
int maxCur = 0, maxSoFar = 0;
for(int i = 1; i < prices.length; i++) {
maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);
maxSoFar = Math.max(maxCur, maxSoFar);
}
return maxSoFar;
}

本文首发于Code & Fun