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【LeetCode】Best Time to Buy and Sell Stock

2024-07-04 00:51:27   224人阅读

Best Time to Buy and Sell Stock

 

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.

 

这题不难，第一反应就是O(n)解决。

关键点就在于，随着prices的遍历，怎样更新gap？

我的做法是，

如果当前遇到的元素比当前max大，就更新max并更新gap

如果当前遇到的元素比当前min小，就重新进行gap的计算

成功AC，上代码

class Solution {
public:
    int maxProfit(vector<int> &prices) 
    {
        if(prices.empty())
            return 0;
        
        int maxGap = 0;
        int curGap = 0;
        int max = prices[0];
        int min = prices[0];
        curGap = max - min;
        
        for(vector<int>::size_type st = 0; st < prices.size(); st ++)
        {
            if(prices[st] > max)
            {
                max = prices[st];
                curGap = max - min;
                if(curGap > maxGap)
                    maxGap = curGap;
            }
            else if(prices[st] < min)
            {
                min = prices[st];
                max = prices[st];
                curGap = max - min;
            }
        }
        return maxGap;
    }
};

 

后来看到Discussion中一种更简洁的做法，

关键点是，如果最大gap对应的max为第j个元素，那么对应的min必是prices[j]之前的最小元素

class Solution 
{
public:
    int maxProfit(vector<int> &prices) 
    {
        if(prices.empty())
            return 0;
        
        int maxGap = 0;
        int curGap = 0;
        int min = prices[0];
        
        for(vector<int>::size_type st = 1; st < prices.size(); st ++)
        {
            if(prices[st] < min)
                min = prices[st];
            
            else
            {
                curGap = prices[st] - min; 
                if(curGap > maxGap)
                    maxGap = curGap;
            }
            
        }
        
        return maxGap;
    }
};


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