LeetCode: -Dynamic Programming-Best Time to Buy and Sell Stock[121]

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.

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

遍历数组，维护最下值、最大值即可

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int min, max;
        if(prices.empty()){
            return 0;
        }else{
            min = prices[0];
            max = 0;
            for(int i = 1; i < n; i++){
                max = prices[i]-min>max?prices[i]-min:max;
                min = prices[i]>min?min:prices[i];
            }
            return max;
        }
    }
};