[LeetCode 121]Best Time to Buy and Sell Stock(动态规划）

题目内容

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 (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.
问题来源

问题简述

在给定的数组中找到升序两数最大差，若数组为降序返回0。

问题思路

普通的二重循环在问题规模达到10000时即会超时，所以只能采取一重循环的动态规划。所以在遍历过程中，维护不变式：min变量储存已遍历部分最小数，max变量储存已遍历部分升序两数最大差。遍历结束即得到正确答案。

代码示例

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

    }
};

时间复杂度O（n） 空间复杂度O（1）