leetcode-121. Best Time to Buy and Sell Stock

千念飞羽

于 2017-01-04 17:03:30 发布

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分类专栏： JAVA leetcode 文章标签： leetcode java

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本文介绍了一种解决LeetCode上121题“Best Time to Buy and Sell Stock”的算法实现。该问题要求从给定的每日股价数组中找出一次买入和卖出的最佳时机以获得最大利润。文章提供了一个Java解决方案，通过维护一个滚动最小值数组和计算最大收益来高效地解决问题。

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leetcode-121. Best Time to Buy and Sell Stock

题目：

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.

题目是找最大收益的一次，方法也很简单，首先需要一个数组去维护当前坐标之前的最小值，然后用一个变量去记录当前值和当前值之前最小值之差就好。当然用两个变量也可以。

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length<=1) return 0;

        int[] min = new int[prices.length];
        min[0] = prices[0];
        int ret = Integer.MIN_VALUE;
        for(int i = 1 ; i < min.length ; i++){
            min[i] = Math.min(min[i-1],prices[i]);
            ret = Math.max(ret,prices[i]-min[i]);
        }
        return ret;
    }
}