121. Best Time to Buy and Sell Stock

Problem

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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

Solution

思路简单到不行，但是…脑子好累，做不动了…

保存三个值：

1. 当前元素之前的最小值；
2. 当前最小利润（局部最优）；
3. 全局最优

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

runtime: 9ms