Leetcode之Best Time to Buy and Sell Stock

标签：leetcode

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.

We solve the problem by examples:

Suppose the input array is: [2, 1, 3, 4, 5, 4]. 
Then, the corresponding diagram is as the following:
             5
            / \
       4   /   4
      / \ /
     3   3
2   /
 \ /
  1

So, what we actually have to do is to find the difference value between the max and the min.

int maxProfit(vector<int> &prices) {
        if(prices.size() == 0)return 0;
        
        int min = prices[0];
        int max = 0;
        for(int i = 0; i < prices.size(); ++i)
        {
            if(prices[i] - min > max)
                max = prices[i] - min;
            
            if(prices[i] < min)
                min = prices[i];
        }
        
        return max;
    }

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Leetcode之Best Time to Buy and Sell Stock

标签：leetcode

原文地址：http://blog.csdn.net/dacxu/article/details/33741495