//The concept here is to negate each number's index as the input is 1 <= a[i] <= n (n = size of array). // Once a value is negated, if it requires to be negated again then it is a duplicate. public List<Integer> findDuplicates(int[] nums) { List<Integer> res = new ...
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 #...
Given an array of integersnumscontainingn + 1integers where each integer is in the range[1, n]inclusive. There is only one duplicate number innums, returnthis duplicate number. Follow-ups: How can we prove that at least one duplicate number must exist innums? Can you solve the problem wi...
classSolution:deffindDuplicate(self, nums:List[int]) ->int:# 如果只有两个元素,第一个元素一定是重复元素iflen(nums) ==2:returnnums[0]# fast每次走两步,slow每次走一步,起始点可以为任意位置fast =0slow =0# python没有do while,所以在循环外写了一遍slow = nums[slow] fast = nums[nums[fast]]...
There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比如取数组为{1,2,3,3,4,5},mid应该是(5+1)/2 = 3,那么,如果小于等于mid的数的个数如果超过了3,那么重复的数字一定出现在l,mid之间,否则出现在mid + 1,r之间。以该...
24. Find Missing Number in an Array of Numbers Between 10 and 20 Write a Python program to find the missing number in a given array of numbers between 10 and 20. Sample Solution-1: Python Code: import array as arr def test(nums): ...
287. Find the Duplicate Number 方法0: 方法1: binary search 方法2: Complexity 易错点 Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one dupl... 查看原文 python3题解 LeetCode剑指 Offer 64. 求1+2+…+n 287. 寻找...
Python Array Exercises, Practice and Solution: Write a Python program to find the first duplicate element in a given array of integers. Return -1 if there are no such elements.
Below is the JavaScript program to find the third maximum number in an array ?Open Compiler const arr = [1, 5, 23, 3, 676, 4, 35, 4, 2]; const findThirdMax = (arr) => { let [first, second, third] = [-Infinity, -Infinity, -Infinity]; for (let el of arr) { if (el ...
Learn how to find the lost element from a duplicated array in JavaScript with our comprehensive guide and example code.