Leetcode 121. Best Time to Buy and Sell Stock 股票买卖 解题报告
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1 解题思想
这道题是用一个数组表示了一个股票不同日期的价值,现在允许我们买一次,卖一次,买肯定要在卖之前。
求可能的最大利润是多少?
这道题只需要直接遍历且:
1、保存已经遍历过的地方的最小售价
2、判断当前位置的售价和当前的最小售价差(当前位置的最大利润)是否超过全局最大利润,若超过了就更新
3、遍历完成后返回全局最大利润
2 原题
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.
3 AC解
public class Solution { /** * 保存一个当前位置以前的,最小的售价, * * 注意题目要求就只是买卖一次,所以在最低价时买,当前位置处比较 * * */ public int maxProfit(int[] prices) { int result=0; if(prices.length<2) return result; int min=prices[0]; for(int i=1;i<prices.length;i++){ result=Math.max(result,prices[i]-min); min=Math.min(min,prices[i]); } return result; }}
0 0
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