daily leetcode - best-time-to-buy-and-sell-stock-ii - !

题目地址

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/

题目描述

Say you have an array for which the ith 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.

思路

这道跟之前那道Best Time to Buy and Sell Stock 买卖股票的最佳时间很类似,但都比较容易解答。这道题由于可以无限次买入和卖出。我们都知道炒股想挣钱当然是低价买入高价抛出,那么这里我们只需要从第二天开始,如果当前价格比之前价格高,则把差值加入利润中,因为我们可以昨天买入,今日卖出,若明日价更高的话,还可以今日买入,明日再抛出。以此类推,遍历完整个数组后即可求得最大利润。代码如下:

C++ 解法:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0, n = prices.size();
        for (int i = 0; i < n - 1; ++i) {
            if (prices[i] < prices[i + 1]) {
                res += prices[i + 1] - prices[i];
            }
        }
        return res;
    }
};

Java 解法:

public class Solution {
    public int maxProfit(int[] prices) {
        int res = 0;
        for (int i = 0; i < prices.length - 1; ++i) {
            if (prices[i] < prices[i + 1]) {
                res += prices[i + 1] - prices[i];
            }
        }
        return res;
    }
}

思路2

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:

best-time-to-buy-and-sell-stock-ii

关键点解析

  • 这类题只要你在心中(或者别的地方)画出上面这种图就很容易解决

代码

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]))

本文参考自:
https://github.com/grandyang/leetcode/ &
https://github.com/azl397985856/leetcode


标题: daily leetcode - best-time-to-buy-and-sell-stock-ii - !
文章作者: lonuslan
文章链接: https://www.lonuslan.com/articles/2020/07/07/1594091102179.html
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