【C++】【LeetCode】121. Best Time to Buy and Sell Stock

题目

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.

Example 1:

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)

Example 2:

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

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

思路

直接粗暴的方法，遍历数组中每个元素，找到该元素之后最大的一个元素，算差值，找到差值最大的一个。

代码

class Solution {public:    int maxProfit(vector<int>& prices) {        int max = 0;        for (int i = 0; i < prices.size(); i++) {            int maxNum = prices[i];            for (int j = i; j < prices.size(); j++) {                maxNum = prices[j] > maxNum ? prices[j] : maxNum;            }            max = (maxNum - prices[i]) > max ? (maxNum - prices[i]) : max;        }        return max;    }};