LeetCode121. Best Time to Buy and Sell Stock

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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 (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: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

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题目的本质是计算数组中两个数的最大差值(右减左)。

主要是维护两个变量。遍历过的元素中的当前最小值min，以及当前的最大差值maxprofit。通过比较num[j]-min和maxprofit来不断的更新maxprofit。

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/solution/ 

https://github.com/abesft/leetcode/blob/master/121BestTimeToBuyAndSellStock/121BestTimeToBuyAndSellStock.cpp 

#include <iostream>
#include<vector>
#include<limits>
using namespace std;


class Solution {
public:
	int maxProfit(vector<int>& prices) {
		if (prices.size() == 0)
			return 0;

		int maxprofit = 0;
		//当前波谷
		int minval = INT_MAX;

		for (int i = 0; i < prices.size(); i++)
		{
			if (prices[i] < minval)
				minval = prices[i];
			else if (maxprofit < prices[i] - minval)
				maxprofit = prices[i] - minval;
		}
		return maxprofit;
	}
};

int main()
{
	Solution sln;
	vector<int> testcase{ 7,1,5,3,6,4 };
	cout << sln.maxProfit(testcase) << endl;
	std::cout << "Hello World!\n";
}

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