一、题目 给定一个整数数组 a,其中1 ≤ a[i] ≤ n (n为数组长度), 其中有些元素出现两次而其他元素出现一次。 找到所有出现两次的元素。 你可以不用到任何额外空间并在O(n)时间复杂度内解决这个问题吗? 示例: 输入: [4,3,2,7,8,2,3,1] 输出: [2,3] 二、题解 解法1:哈希表 遍历给定数组时,可以查看该元素是否存在于哈希表
再遍历,如果两个相邻元素相等,则会有重复元素。存入list。 publicList<Integer> findDuplicates(int[] nums) { List<Integer> list =newArrayList<>(); Arrays.sort(nums);for(inti=1;i<nums.length;i++) {if(nums[i]==nums[i-1]) { list.add(nums[i]); } }returnlist; } 2. O(N): 题目已知...
442. Find All Duplicates in an ArrayMedium Topics Companies Given an integer array nums of length n where all the integers of nums are in the range [1, n] and each integer appears at most twice, return an array of all the integers that appears twice. You must write an algorithm that ...
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 extra space and in O(n) runtime? Example: Input: [4,3,2,7,8,2,3,1] Output:...
https://leetcode.com/problems/find-all-duplicates-in-an-array/ 典型的数组中的重复数。这次是通过跳转法,一个个跳转排查的。因为查过的不会重复处理,所以复杂度也是O(n)。 后面发现了别人一个更好的做法。。。如下: public class Solution {
Can you solve this real interview question? Remove Duplicates from Sorted Array - Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place [https://en.wikipedia.org/wiki/In-place_algorithm] such that each unique element
26. 删除有序数组中的重复项 - 给你一个 非严格递增排列 的数组 nums ,请你 原地 [http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95] 删除重复出现的元素,使每个元素 只出现一次 ,返回删除后数组的新长度。元素的 相对顺序 应该保持 一致 。然后返回 n
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 extra space and in O(n) runtime?
Do not allocate extra space for another array, you must do this bymodifying the input arrayin-placewith O(1) extra memory. 增加一个count来计数,从左至右遍历,遇到相同的直接del,否则count += 1 classSolution:defremoveDuplicates(self,nums):""" ...
Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. ...