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> res = new ArrayList(); for(int i = 0; i < nums.length; i++){ int val = Math.abs(nums[i]) - 1; if(nums[val] > 0) nums[val] = -nums[val]; } for(int i = 0; i < nums.length; i++){ if(nums[i] > 0){ res.add(i+1); } } return res; } }
利用 1 ≤ a[i] ≤ n 这个条件
然后val 从来没等于过4,5.导致更新后的array 4,5还是正的,然后说明了array里面5,6没有出现过