LeetCode刷题【Array】 Best Time to Buy and Sell Stock

最新推荐文章于 2024-01-03 12:56:44 发布

SimonCode

最新推荐文章于 2024-01-03 12:56:44 发布

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分类专栏： LeetCode 文章标签： Array leetcode

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本文探讨了给定股票价格数组情况下如何通过一次交易获得最大利润的问题，并提供了三种不同的算法实现方式，包括暴力求解、一次遍历及使用Kadane's Algorithm算法。

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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.

解决方法一： 对于比较极端的案例，会Time Limit Exceeded，后面考虑优化;

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices==null) return 0;
        int profit=0;
        int l[] = new int[prices.length];
        for(int i=0;i<prices.length;i++){
            for(int j=i;j>=0;j--){
                l[j]= (prices[i]-prices[j])>l[j] ? (prices[i]-prices[j]):l[j];
                if(l[j]>profit)
                    profit=l[j];
            }
        }
        return profit;
    }
}

解决方法二： Runtime: 2 ms

public class Solution {
    public int maxProfit(int[] prices) {
        if(prices==null) return 0;
        int profit=0;
        int minbuy=2147483647;
        for(int i=0;i<prices.length;i++){
            minbuy=prices[i]<minbuy ? prices[i]:minbuy;
            profit=(prices[i]-minbuy)>profit ? (prices[i]-minbuy):profit;
        }
        return profit;
    }
}

解决方法三：Kadane's Algorithm 算法, Runtime: 4 ms

public class Solution {
    public int maxProfit(int[] prices) {
        int maxCur = 0, maxSoFar = 0;
        for(int i = 1; i < prices.length; i++) {
            maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);
            maxSoFar = Math.max(maxCur, maxSoFar);
        }
        return maxSoFar;
    }
}

参考：

【1】https://leetcode.com/

【2】https://en.wikipedia.org/wiki/Maximum_subarray_problem