LeetCode 121: Best Time to Buy and Sell Stock

原创于 2013-08-22 13:42:41 发布 · 675 阅读

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本文介绍了一种解决股票交易问题的算法，目标是在只完成一次买卖操作的情况下获取最大利润。通过遍历股价数组，算法能够高效地找出最优买卖时机。

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Difficulty: 2

Frequency: 1

Problem:

Say you have an array for which the i^th 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.

Solution:

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int i_max_index = 0, i_min_index = 0;
        int i_max_profit = 0;
        for (int i = 1; i<prices.size(); i++)
        {
            if (prices[i]<prices[i_min_index])
            {
                i_min_index = i_max_index = i;
            }
            else if (prices[i]>=prices[i_max_index])
            {
                i_max_index = i;
                if (i_max_profit<prices[i_max_index] - prices[i_min_index])
                    i_max_profit = prices[i_max_index] - prices[i_min_index];
            }
        }
        return i_max_profit;
    }
};

Note:

I've seen someone solve this problem using DP. Its time complexity is O(nlogn), which is not optimal. My solution's time complexity is O(n);