买卖股票的最佳时机 Best Time to Buy and Sell Stock

题目

给定一个数组,它的第 i 个元素是一支给定股票第 i 天的价格。如果你最多只允许完成一笔交易(即买入和卖出一支股票),设计一个算法来计算你所能获取的最大利润

以下为详细说明:

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: 5

Explanation: 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. Example 2:

Input: [7,6,4,3,1]
Output: 0

Explanation: In this case, no transaction is done, i.e. max profit = 0.

思路&题解

  1. 每一次买入之后,剩下的时间循环计算所有卖出可能性,算出最大利润;
  2. 冲第一天开始,循环计算买入价格,结合步骤一得到每次不同买入后可能得到的最大利润,取大值。
public int maxProfit(int[] prices) {
    int maxProfit=0;
    for(int i = 0; i < prices.length; i++){
        int buy = prices[i];
        int sell = buy;
        for(int j = i+1; j < prices.length; j++){
            if(prices[j] > sell) {
                sell = prices[j];
            }
        }
        maxProfit = Math.max(sell-buy, maxProfit);
    }
    return maxProfit;
}

知识点分析

虽然前面的计算通过了提交,但它不是最优算法。下面介绍下另一种算法:

121_profit_graph

核心在于一次循环找出最低价和最大利润。由于减少了一重循环他的时间复杂度只有O(n)

public class Solution {
    public int maxProfit(int prices[]) {
        int minprice = Integer.MAX_VALUE;
        int maxprofit = 0;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < minprice)
                minprice = prices[i];
            else if (prices[i] - minprice > maxprofit)
                maxprofit = prices[i] - minprice;
        }
        return maxprofit;
    }
}
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