【leetcode系列】122. 买卖股票的最佳时机 II

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。

/*
 * @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;
};

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.