https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
给定一个数组,它的第 i 个元素是一支给定股票第 i 天的价格。
如果你最多只允许完成一笔交易(即买入和卖出一支股票),设计一个算法来计算你所能获取的最大利润。
注意你不能在买入股票前卖出股票。
示例 1:
输入: [7,1,5,3,6,4]
输出: 5
解释: 在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 = 6)的时候卖出,最大利润 = 6-1 = 5 。
注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格。
示例 2:
输入: [7,6,4,3,1]
输出: 0
解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock
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由于我们是想获取到最大的利润,我们的策略应该是低点买入,高点卖出。
由于题目对于交易次数有限制,只能交易一次,因此问题的本质其实就是求波峰浪谷的差值的最大值。
用图表示的话就是这样:
语言支持:JS,Python,C++
JS Code:
/*
* @lc app=leetcode id=121 lang=javascript
*
* [121] Best Time to Buy and Sell Stock
*
* https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
*
* algorithms
* Easy (46.34%)
* Total Accepted: 480.5K
* Total Submissions: 1M
* Testcase Example: '[7,1,5,3,6,4]'
*
* Say you have an array for which the i^th 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.
*
*
*/
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
let min = prices[0];
let profit = 0;
// 7 1 5 3 6 4
for(let i = 1; i < prices.length; i++) {
if (prices[i] > prices[i -1]) {
profit = Math.max(profit, prices[i] - min);
} else {
min = Math.min(min, prices[i]);;
}
}
return profit;
};
Python Code:
# 应用Kadane's algorithms
class Solution:
def maxProfit(self, prices: 'List[int]') -> int:
"""
step by step
"""
# error case
if len(prices) < 1:
return 0
# caluate the daily gains, break into a subarray problem
gains = [prices[i]-prices[i-1] for i in range(1, len(prices))]
loc_max = global_max = 0 #not gains[0] in case of negative
for i in range(len(gains)):
loc_max = max(loc_max + gains[i], gains[i])
if loc_max > global_max:
global_max = loc_max
"""
Runtime: 48 ms, faster than 34.50% of Python3 online submissions for Best Time to Buy and Sell Stock.
Memory Usage: 14.1 MB, less than 10.26% of Python3 online submissions for Best Time to Buy and Sell Stock.
"""
C++ Code:
/**
* 系统上C++的测试用例中的输入有[],因此需要加一个判断
*/
class Solution {
public:
int maxProfit(vector<int>& prices) {
if (prices.empty()) return 0;
auto min = prices[0];
auto profit = 0;
for (auto i = 1; i < prices.size(); ++i) {
if (prices[i] > prices[i -1]) {
profit = max(profit, prices[i] - min);
} else {
min = std::min(min, prices[i]);;
}
}
return profit;
}
};