第四周leetcode题

最新推荐文章于 2024-03-14 22:15:58 发布

原创最新推荐文章于 2024-03-14 22:15:58 发布 · 267 阅读

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本文介绍了一种寻找股票交易最大利润的算法实现方法，通过优化循环结构减少比较次数，提高效率。

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Description:

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

第一次做的时候想当然地用了两层循环：

class Solution {
public:
int maxProfit(vector<int>& prices) {

int max=0;
for(int i=0;i<prices.size();i++)
{
for(int j=i+1;j<prices.size();j++)
{
if((prices[j]-prices[i])>max)max=prices[j]-prices[i];
}
}

return max;

}
};

结果当然就是超时。于是改进了一下，在一层循环中改变max的值来减少比较次数。另外还要注意如果参数的向量为空的话要返回0.

代码如下：

class Solution {
public:
int maxProfit(vector<int>& prices) {

if(prices.size()==0)return 0;

int max=0;
int min=prices[0];

for(int i=1;i<prices.size();i++)
{
   if(prices[i]<min)min=prices[i];
   if(max<prices[i]-min)max=prices[i]-min;
}

return max;

}
};