448. Find All Numbers Disappeared in an Array
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448. Find All Numbers Disappeared in an Array
题目
Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
Example:
Input:[4,3,2,7,8,2,3,1]Output:[5,6]
翻译
给定整数的数组,其中1≤A [1]≤ Ñ(Ñ =数组的大小),一些元件出现两次以及其他出现一次。
找到不出现在这个数组中的[1,n ]包含的所有元素。
你可以做到这一点无需额外的空间和O(ñ)运行?您可以假设返回的列表不计入额外的空间。
例:
输入:
[4,3,2,7,8,2,3,1]
输出:
[5,6]
解题思路
public class Solution { public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> list = new ArrayList<Integer>(); for(int i=0;i<nums.length;i++){ nums[Math.abs(nums[i])-1] = -Math.abs(nums[Math.abs(nums[i])-1]); } for(int i=0;i<nums.length;i++){ if(nums[i]>0) list.add(i+1); } return list; }}
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