LeetCode Array Easy121. Best Time to Buy and Sell Stock

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [,,,,,]
Output:
Explanation: Buy on day  (price = ) and sell on day  (price = ), profit = - = .
             Not - = , as selling price needs to be larger than buying price.

Example 2:

Input: [,,,,]
Output:
Explanation: In this case, no transaction is done, i.e. max profit = .

题目描述： 给一个数组，每一个元素都代表当前索引的价格。你只有一次买进和卖出的机会，计算买进最低点和卖出最高点。并返回最大差价

思路：首先判断当前点是否可以是买进点。及当前点是否比下一点小。

2，在当前点是买进点的时候，从当前点向尾部遍历，判断当前最大差价点。

3，不断进行比较，找到整体的出现最大差价时的买进点和卖出点

代码如下：

public int MaxProfit(int[] prices)
        {
            int inIndex = -, outIndex = -;//买进点， 卖出点
            int subMax = int.MinValue;//存储最大差价
            for (int i = ; i < prices.Length -; i++)
            {
                if(prices[i] < prices[i + ])//如果当前值比下一个值小，说明此时买进可以有盈利
                {
                    int tempMax = ;
                    int maxIndex = -;
                    for (int j = i; j < prices.Length; j++)//从买进处遍历，判断最大差价点的索引
                    {
                        int temp = prices[j] - prices[i];
                        if(temp > tempMax)
                        {
                            tempMax = temp;
                            maxIndex = j;
                        }
                    }
                    if(tempMax > subMax)//更新买进点、卖出点
                    {
                        subMax = tempMax;
                        inIndex = i;
                        outIndex = maxIndex;
                    }
                }
            }
            return subMax >  ? prices[outIndex] - prices[inIndex] : ;

        }