LeetCode_Best Time to Buy and Sell Stock

最新推荐文章于 2020-04-25 11:09:15 发布

shengno1

最新推荐文章于 2020-04-25 11:09:15 发布

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分类专栏： Something_Interesting Leetcode C++ xOJ_Coding 文章标签： C++ leetcode 算法题

本文链接：https://blog.youkuaiyun.com/sheng_ai/article/details/50119161

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一.题目

Best Time to Buy and Sell Stock

My Submissions

Total Accepted: 75747 Total Submissions: 221316 Difficulty: Medium

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.

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二.解题技巧

将每一天的股票价格减去前一天的股票价格，就可以将问题转化为求数组的最大子段和的问题，这样就可以用求解最大子段和的方法来解答。

三.实现代码

#include <iostream>
#include <vector>
#include <algorithm>


using namespace std;

class Solution
{
private:
int FindGreatestSumOfSubArray(vector<int>::const_iterator Begin,
                              vector<int>::const_iterator End)
    {
        // empty
        if (Begin == End)
        {
            return 0;
        }

        vector<int>::const_iterator Ite = Begin;
        int TmpResult_1 = 0;
        int TmpResult_2 = *Ite;
        int Sum = 0;
        while(Ite != End)
        {
            if (*Ite < 0)
            {
                if (Sum > TmpResult_1)
                {
                    TmpResult_1 = Sum;
                }
            }

            Sum += *Ite;
            if (Sum < 0)
            {
                Sum = 0;
            }

            if (*Ite > TmpResult_2)
            {
                TmpResult_2 = *Ite;
            }

            ++Ite;
        }

        if (TmpResult_2 < 0)
        {
            return 0;
        }

        if (Sum > TmpResult_1)
        {
            TmpResult_1 = Sum;
        }

        return TmpResult_1;
    }

public:
    int maxProfit(vector<int>& prices)
    {
        if (prices.empty())
        {
            return 0;
        }


        if (prices.size() == 1)
        {
            return 0;
        }

        for (vector<int>::iterator Ite = prices.begin(); Ite != prices.end() - 1;
             ++Ite)
        {
            *Ite = *(Ite + 1) - *Ite;
        }

        return FindGreatestSumOfSubArray(prices.begin(), prices.end() - 1);
    }

};