leetcode448~Find All Numbers Disappeared in an Array

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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

Input:
[4,3,2,7,8,2,3,1]Output:
[5,6]

public  List<Integer> findDisappearedNumbers3(int[] nums) {List<Integer> res = new ArrayList<Integer>();for(int i=0;i<nums.length;i++) {//如果是重复的数，则只会标记一个，变为相反数//将数放到对应的索引位置上 如位置i上放i+1int val = Math.abs(nums[i])-1;if(nums[val]>0) {//标记为负数nums[val] = -nums[val];}}//经过上一步的标记，正常出现的数都变负了for(int i=0;i<nums.length;i++) {if(nums[i]>0) {res.add(i+1);}}return res;}public  List<Integer> findDisappearedNumbers(int[] nums) {List<Integer> res = new ArrayList<Integer>();int len = nums.length;for(int i=0;i<len;i++) {nums[(nums[i]-1)%len] += len;}for(int i=0;i<len;i++) {if(nums[i]<=len) {res.add(i+1);}}return res;}

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