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).
分析:
// 4 7 8 2 8
最大利润很明显是 (8 - 4) + (8 - 2) = 10
就因为这个式子让我想复杂了:首先要找到一个极小值4,然后找到极大值8;然后找到极小值2,然后找到极大值8;balabala……
其实换一种思路,(7 - 4) + (8 - 7) + (8 - 2)
区别在于,直接将后一个数减前一个数就好了呀,只不过如果后一个数比前一个数小的话就不考虑而已。
ac代码:
class Solution {
public:
int maxProfit(vector<int>& prices) {
int i,n=prices.size(),sum=0;
for(i=1;i<n;i++)
{
if(prices[i]-prices[i-1]>0)
sum+=prices[i]-prices[i-1];
}
return sum;
}
};
本文介绍了一种计算股票交易最大利润的算法。该算法允许进行多次买卖操作,但每次卖出后才能再次购买。通过遍历股价数组并计算相邻两天价格之差来确定利润,若差值为正则计入总利润。
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