Write a Python program to find the first duplicate element in a given array of integers. Return -1 if there are no such elements.Sample Solution:Python Code :def find_first_duplicate(nums): num_set = set() no_duplicate = -1 for i in range(len(nums)): if nums[i] in num_set: re...
Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without ex...[数组]442. Find All Duplicates in an Array 标签(空格分隔): 数组 leetcode 刷...
Learn how to find the lost element from a duplicated array in JavaScript with our comprehensive guide and example code.
This post will discuss how to find the duplicate elements in a list in C#... The idea is to use the Enumerable.GroupBy() method to group the elements based on their value, then filter out the groups that appear more than once, and retrieve the duplicates
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LeetCode 287. Find the Duplicate Number 暴力解法 时间 O(nlog(n)),空间O(n),按题目中Note“只用O(1)的空间”,照理是过不了的,但是可能判题并没有卡空间复杂度,所以也能AC。 双指针判断环 时间O(n),空间O(1),思路十分巧妙,但是使用
Python代码:class Solution(object): def findDuplicate(self, paths): """ :type paths: List[str] :rtype: List[List[str]] """ filemap = collections.defaultdict(list) for path in paths: roads = path.split() directory, files = roads[0], roads[1:] for file in files: file_s = file...
The command adds data to the new table, including several duplicate entry combinations. Step 2: Find the Duplicates in MySQL To identify duplicates in MySQL, use queries that locate entries that appear multiple times. Depending on the use case and data complexity, there are several ways to fin...
Find fixed point in an array: In this tutorial, we will learn how to find a fixed point in a given sorted array of distinct elements using both linear search & binary search. If there is no fixed point, then return -1.
代码(Python3) class Solution: def findDuplicate(self, nums: List[int]) -> int: # 二分区间左边界,初始化为 1 l: int = 1 # 二分区间右边界,初始化为 n r: int = len(nums) - 1 # 当前区间不为空时,继续二分 while l <= r: # 计算区间中点 mid mid: int = (l + r) >> 1 #...