标签:排序 pre time 存储 测试 and amp 因此 you
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]
class Solution { public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> list = new ArrayList<>(); if(nums.length <= 1) return list; for(int i = 0; i < nums.length; i++){ while(nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]){ int temp = nums[i]; nums[i] = nums[nums[i] - 1]; nums[temp - 1] = temp; } } for(int i = 0; i < nums.length; i++){ if(nums[i] != i + 1){ list.add(i + 1); } } return list; } }
leetcode448-Find All Numbers Disappeared in an Array
标签:排序 pre time 存储 测试 and amp 因此 you
原文地址:https://www.cnblogs.com/clairexxx/p/10531835.html