LeetCode--No.121--Best Time to Buy and Sell Stock

最新推荐文章于 2022-02-07 21:21:56 发布

sophia_tone2w

最新推荐文章于 2022-02-07 21:21:56 发布

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分类专栏： LeetCode--Easy

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本文介绍了一种寻找股票交易中最大利润的算法。该算法能在O(n)的时间复杂度内找到给定价格数组中能够获得的最大利润，同时考虑了买入价格必须小于卖出价格的实际约束条件。

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.

方法：

时间复杂度O(n^2)的就略过了，比较简单。

方法二：

时间复杂度O(n),空间复杂度O(n), 其实和求子数列的最大和，几乎是一样的。

public class Solution {
    public int maxProfit(int[] prices) {
        int max = 0;
        int min = 0;
        if (prices.length <= 1)
            return 0;
        else if(prices.length == 2){
            if (prices[1] - prices[0] > 0)
                return prices[1] - prices[0];
            else
                return 0;
        }
        int[] sum = new int[prices.length - 1];
        sum[0] = prices[1] - prices[0];
        int maxs = sum[0];
        for(int i = 1; i < prices.length-1; i++){
            sum[i] = Math.max(prices[i+1] - prices[i], sum[i-1] + prices[i+1] - prices[i]);
            maxs = Math.max(maxs, sum[i]);
        }
        if (maxs < 0)
            return 0;
        else
            return maxs;
    }
}

为什么就不能简单一点：？！！

感觉受到了一万点伤害。

我这个智商真的还要继续刷下去么？

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 0)
            return 0;
        int maxPrice = prices[prices.length-1];
        int ans = 0;
        for(int i = prices.length - 1; i>= 0; i--){
            maxPrice = Math.max(maxPrice, prices[i]);
            ans = Math.max(ans, maxPrice - prices[i]);
        }
        return ans;
    }
}