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.

題目的本質是計算陣列中兩個數的最大差值(右減左)。

主要是維護兩個變數。遍歷過的元素中的當前最小值min，以及當前的最大差值maxprofit。通過比較num[j]-min和maxprofit來不斷的更新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;
		//當前波谷
		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";
}