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.

题目思路

• 看示例可知，只要后一天股票价格比前一天大，就卖出

代码 C++

• 思路一、

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size() < 2){
            return 0;
        }
        
        int profit = 0;
        
        for(int i=1; i < prices.size(); i++){
            profit += (prices[i]-prices[i-1] > 0? prices[i]-prices[i-1] : 0);
        }
        
        return profit;
    }
};

• 思路二、

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size() < 2){
            return 0;
        }
        
        int profit = 0;
        
        for(int i=1; i < prices.size(); i++){
            if(prices[i]-prices[i-1] > 0){
                profit += prices[i]-prices[i-1];
            }
        }
        
        return profit;
    }
};

总结展望