leetcode - [122] Best Time to Buy and Sell Stock II

2020/02/16

题目描述

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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

有一个数组,其中索引代表是哪一天,值代表当天的股票价格,求其最大的利润。

举两个例子:

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

解法1

我的想法是这样的:

首先,从第二个数值开始遍历,若其大于前一个数值,则累加其差,否则跳过。

@Test
public void question122(){
    int[] prices = {7, 1, 5, 3, 6, 4};
    int ret = ArrayProblems.p122(prices);
    Assert.assertEquals(ret, 7);
}

public class ArrayProblems {
    public static int p122(int[] prices) {
        int amount = 0;
        for (int i = 1; i < prices.length; i++){
            if (prices[i] > prices[i - 1]){
                amount += (prices[i] - prices[i - 1]);
            }
        }
        return amount;
    }
}

优化1

翻了下讨论区,有种更简洁的写法:

public class ArrayProblems {
    public static int p122(int[] prices) {
        int profit = 0;
        for (int i = 1; i < prices.length; i++){
            profit += Math.max(0, prices[i] - prices[i - 1]);
        }
        return profit;
    }
}

解法2

再来看一种解法,如下图所示:

public static int p122UsingResetPoint(int[] prices){
    if(prices.length == 0) return 0;
    int profit = 0, low = prices[0], max = 0;
    for(int i = 1; i < prices.length; i++){
        if(prices[i] < prices[i - 1]){
            //reset point
            profit += max;
            max = 0;
            low = prices[i];
        }else {
            max = Math.max(max, prices[i] - low);
        }
    }
    profit += max;
    return profit;
}

Reference


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