LeetCode121 Best Time to Buy and Sell Stock

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题目要求:
121. Best Time to Buy and Sell Stock QuestionEditorial Solution My Submissions

Total Accepted: 139457
Total Submissions: 364702
Difficulty: Easy
Contributors: Admin
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.

ac的代码(比较简单)

public class Solution {public int maxProfit(int[] prices) {if(prices.length <= 1) return 0;int min = prices[0];int result = 0;for(int i = 1; i < prices.length; i++) {min = Math.min(prices[i], min);result = Math.max(result, prices[i] - min);}return result;}
}

第一次写的代码,没有接受(超时,使用的是最简单的循环的方式)

package com.syy.leetcode121;import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;//该方法时间超时public class solution_new {public int maxProfit(int[] prices) {if(prices.length <=1) return 0;List<Integer> list = new ArrayList<Integer>();for(int i = 0; i < prices.length; i++) {for(int j = i; j < prices.length; j++) {if(prices[i] < prices[j]) list.add(prices[j] - prices[i]);}}Collections.sort(list);return list.size()>0 ? list.get(list.size()-1) : 0;}
}

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