《leetCode》：Best Time to Buy and Sell Stock II

题目

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 (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

思路

贪婪算法 :判断相邻的两个元素之差是否是大于零，如果大于零，则可以构成一次交易；
如果相邻的两个元素之差小于零，则可以理解为当天买当天卖，利润为零。综上所述：统计所有相邻的元素大于零的之和即为最大收益

实现代码如下：

int maxProfit(int* prices, int pricesSize) {    if(prices==NULL||pricesSize<1){        return 0;    }    int profit=0;    for(int i=1;i<pricesSize;i++){        profit+=(prices[i]-prices[i-1])>0?(prices[i]-prices[i-1]):0;    }    return profit;  }