[LeetCode]122. Best Time to Buy and Sell Stock II 解题报告(C++)

[LeetCode]122. Best Time to Buy and Sell Stock II 解题报告(C++)

题目描述

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.

题目大意

• day i 的价钱是 arr[i]. 可以通过无限次的交易.使得收益最大.
• 注意: 二次购买必须把第一次购买的卖出去.

解题思路

方法1:

• 由于没有限制买卖次数. 所以收益可以叠加.
• 就是要找到前面价格比后面低的位置.
• 然后一直叠加收益即可.

代码实现:

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

        }
        return res;
    }
};

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小结

• 收益叠加的方法! 真的nice! 因为不限制交易次数!