121. 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.

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.

第一种思路：最容易想到的方法，时间复杂度O(n^2);

  int maxProfit(vector<int>& prices) {
        int max=0;
        int size=(int)prices.size();
        for(int i=0; i<size;i++)
        {
            int base=prices[i];
            for(int j=i+1; j<size; j++)
                max=(max<(prices[j]-base))?prices[j]-base:max;
        }
        return max;
    }

第二种方法：时间复杂度O(n)。保存当前元素前的最小值，求它们之间的差值，若大于变量max, 则修改max的值。

  int maxProfit(vector<int>& prices) {
       int min=0x7FFFFFFF, max=0;
       for(auto val:prices)
       {
           min=fmin(min,val);
           max=fmax(max, val-min);
       }
        return max;
    }