题目地址:Best Time to Buy and Sell Stock - LeetCode
买股票系列题目:
- LeetCode 121. Best Time to Buy and Sell Stock–Java,Python,C++解法
- LeetCode 122. Best Time to Buy and Sell Stock II–贪心–Java,C++,Python解法
- LeetCode 123. Best Time to Buy and Sell Stock III–Python解法–动态规划–数学题
- LeetCode 309. Best Time to Buy and Sell Stock with Cooldown–Java解法-卖股票系列题目
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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
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.
Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
这道题目遍历一遍即可。 Python解法如下:
class Solution:
def maxProfit(self, prices: List[int]) -> int:
maxprofit=0
minprice=-1
for i in prices:
if minprice==-1:
minprice=i
elif i<minprice:
minprice=i
elif i-minprice>maxprofit:
maxprofit=i-minprice
return maxprofit
Java解法如下:
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;
}
}
C++解法如下:
class Solution {
public:
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;
}
};
文档信息
- 本文作者:last2win
- 本文链接:https://last2win.com/2020/01/22/LeetCode-121-Best-Time-to-Buy-and-Sell-Stock/
- 版权声明:自由转载-非商用-非衍生-保持署名(创意共享3.0许可证)