leetcode刷题记录-121. Best Time to Buy and Sell Stock

leetcode刷题记录-121. Best Time to Buy and Sell Stock

1.题目要求

　　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.

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2.解题思路

• 思路一
  　　依次遍历数组中的每个数，然后计算后面的数与它的差，取最大的值返回即可。
  　　代码如下：

    int maxProfit(vector<int>& prices) 
    {
        int output=0;
        int i,j;
        if(prices.size()==0)
        {
            return 0;
        }
        for(i=0;i1;i++)
        {
            for(j=i+1;jif(prices[j]-prices[i]>output)
                    output=prices[j]-prices[i];
            }

        }

        return output;
    }

　　提交的时候报错，时间复杂度太高，输入的n越大，那么耗时越长。。。所以不可取。于是进行改进，只需要遍历一次即可，见思路二。

• 思路二
  　　依次遍历数组，找出最小值，那么需要返回的值一定是当前的值与最小值的差（要求大于等于0）取最大的那个值，实现代码如下：

    int maxProfit(vector<int>& prices) 
    {
        int output=0;
        int max_price,min_price;
        if(prices.size()==0)
        {
            return 0;
        }
        min_price=prices[0];
        max_price=0;
        for(i=0;ireturn output;
    }

　　参考评论区的改进。。另外就是循环的时候其实每次只需要计算min_price和max_price中的一个就行了。

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3.总结

　　总的来说，这道题目还是比较简单的，细节地方注意一下即可。