leetcode121 Best Time to Buy and Sell Stock

xiaoxiao2021-02-28  7

问题描述

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: 5Explanation: 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.Input: [7,6,4,3,1]Output: 0Explanation: In this case, no transaction is done, i.e. max profit = 0.

Example 2:

中文题意

给定一个数组,其中存放每天的股票价格,在价格低的时候买进,价格高的时候抛出,才能获利,求最大的利益(只能进行一次完整的买进-抛出交易)。

方法一

使用暴力遍历法,即求数组中任意两个元素的差值,然后找到这些差值的最大值。

class Solution { public int maxProfit(int[] prices) { int max = 0; for(int i = 0; i < prices.length - 1; i++){ for(int k = i+1; k<prices.length;k++){ if(prices[i] - prices[k] >=0){ continue; } max = max > prices[k] - prices[i] ? max : prices[k] - prices[i]; } } return max; } }

在测试中,容易出现Time Limit Exceeded现象,时间复杂度为O(n2)。

方法二

方法一在许多问题中都可以使用,方法二针对本问题,做一个优化。

方法一造成超时,是因为对任意的两个数字都做求差值计算,现在的优化是:获得位置i上的元素,i位之前存在的最小值为curmin,只需要计算第i位的元素和curmin的差值即可。同时profitmax变量存储整个数组范围内的最大利益值。

class Solution { public int maxProfit(int[] prices) { if(prices.length ==0) return 0; int curmin = prices[0];//记录当前位置之前的最小元素 int curmax = 0;//当前位置的元素和当前最小元素的差值 int profitmax = 0;//整个数组范围内的最大利益值 for(int i = 1; i < prices.length; i++){ curmax = prices[i] - curmin; if(prices[i] < curmin) curmin = prices[i]; if(curmax > profitmax) profitmax = curmax; } return profitmax; } }
转载请注明原文地址: https://www.6miu.com/read-1900374.html

最新回复(0)