[4]169. Majority Element/[4]121. Best Time to Buy and Sell Stock(Java)

最新推荐文章于 2025-07-20 10:45:54 发布

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最新推荐文章于 2025-07-20 10:45:54 发布

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分类专栏： leetcode Array

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本文介绍两种寻找数组中多数元素的方法，并提供一种算法用于确定股票交易的最佳时机以获得最大利润。第一种方法使用O(n)的时间复杂度和O(1)的空间复杂度来找到出现次数超过一半的元素；第二种方法通过排序找到中间位置的元素作为多数元素。此外，还提出了一种算法来解决股票买卖问题，帮助找到买入和卖出股票的最佳时间。

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Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

in Java(1) : O(n) time O(1) space

class Solution {
    public int majorityElement(int[] nums) {
        int major = nums[0], count = 1;
        for (int i = 1; i < nums.length; i ++) {
            if (nums[i] == major) count ++;
            else count --;
            if (count == 0) {
                major = nums[i];
                count ++;
            }
        }
        return major;
    }
}

in Java(2):

public class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length / 2];
    }
}

Best Time to Buy and Sell Stock(Java)

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.

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