448. Find All Numbers Disappeared in an Array

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]

Answer:

public class Solution {    public List<Integer> findDisappearedNumbers(int[] nums) {        List<Integer> res = new ArrayList<Integer>();        if(nums.length==0||nums==null){            return res;        }        Arrays.sort(nums);        for(int i=0;i<nums.length;i++){            if(nums[i] != i+1){                if(Arrays.binarySearch(nums, i+1)<0){                    res.add(i+1);                }            }        }        return res;    }} 

思路： 网上看的答案理解不太能理解，研究了一套自己看的懂的代码，先对数组进行排序，然后循环判断数组该位置对应的数值是否为数组按序的值，如果不是就添加进数组去。