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。
设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易(多次买卖一支股票)。
注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。
示例 1:
输入: [7,1,5,3,6,4]
输出: 7
解释: 在第 2 天(股票价格 = 1)的时候买入,在第 3 天(股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。
随后,在第 4 天(股票价格 = 3)的时候买入,在第 5 天(股票价格 = 6)的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。
示例 2:
输入: [1,2,3,4,5]
输出: 4
解释: 在第 1 天(股票价格 = 1)的时候买入,在第 5 天 (股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。
注意你不能在第 1 天和第 2 天接连购买股票,之后再将它们卖出。
因为这样属于同时参与了多笔交易,你必须在再次购买前出售掉之前的股票。
示例 3:
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-ii
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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没有限制,因此只要能够赚钱的机会我们都不应该放过。
如下图,我们只需要求出加粗部分的总和即可
用图表示的话就是这样:
语言支持:JS,Python
JS Code:
/*
* @lc app=leetcode id=122 lang=javascript
*
* [122] Best Time to Buy and Sell Stock II
*
* https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
*
* algorithms
* Easy (50.99%)
* Total Accepted: 315.5K
* Total Submissions: 610.9K
* 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.
*
* 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.
*
*/
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
let profit = 0;
for(let i = 1; i < prices.length; i++) {
if (prices[i] > prices[i -1]) {
profit = profit + prices[i] - prices[i - 1];
}
}
return profit;
};
Python Code:
class Solution:
def maxProfit(self, prices: 'List[int]') -> int:
gains = [prices[i] - prices[i-1] for i in range(1, len(prices))
if prices[i] - prices[i-1] > 0]
return sum(gains)
print(Solution().maxProfit([7, 1, 5, 3, 6, 4]))
#评论区里都讲这是一道开玩笑的送分题.