【Leetcode】Best Time to Buy and Sell Stock III

最新推荐文章于 2025-08-11 14:43:21 发布

yyz20002008

最新推荐文章于 2025-08-11 14:43:21 发布

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文章标签： java leetcode array

本文介绍了一种算法，用于计算给定股票价格数组中最多进行两次买卖交易所能获得的最大利润。该算法首先通过一次遍历找到第一次买入和卖出的最佳时机，然后再次遍历找出第二次交易的最佳时机。最终通过组合两次交易的利润得出最大收益。

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Best Time to Buy and Sell Stock III

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Java:

http://blog.youkuaiyun.com/u013027996/article/details/19414967

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0){
            return 0;
        }
        int len = prices.length;
        int maxProfit = 0;
        int min = prices[0];
        int arrayA[] = new int[len];
        for(int i = 1; i < len; i++){
            arrayA[i] = prices[i] - min;
            arrayA[i] = arrayA[i] > arrayA[i-1] ? arrayA[i] : arrayA[i-1];
            if(prices[i] < min){
                min = prices[i];
            }
        }
        int max = prices[len-1];
        int arrayB[] = new int[len];
        for(int i = len-2; i >= 0; i--){
            arrayB[i] = max - prices[i];
            arrayB[i] = arrayB[i] > arrayB[i+1] ? arrayB[i] : arrayB[i+1];
            if(prices[i] > max){
                max = prices[i];
            }
        }
        for(int i = 0; i < len; i++){
            int tempValue = arrayA[i] + arrayB[i];
            maxProfit = maxProfit > tempValue ? maxProfit : tempValue;
        }
        return maxProfit;
    }
}