448. Find All Numbers Disappeared in an Array

题目摘要
给定一个整形数组，其中1≤a[i]≤n(n为数组长度)，一些元素出现两次一些元素只出现一次，找到1~n中所有没有出现的元素。请用O(n)的时间复杂度，并且不使用额外的空间。

解法
1. 遍历数组元素，将n[n[i] - 1]置负，再遍历一次，如果n[i]不为负，则i + 1没有出现。
2. 遍历数组元素，对于每一个元素，如果n[i] != i + 1 && n[n[i] - 1] != n[i]则调换n[i]和n[n[i] - 1]直到上述条件成立，再遍历一次，如果n[i] != i + 1则i + 1没有出现

注意

可问问题

原题
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]

Solution#1

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

Solution#2

public class Solution {    public List<Integer> findDisappearedNumbers(int[] nums) {        List list = new ArrayList<Integer>();        if (nums == null || nums.length == 0) return list;        for (int i = 0; i < nums.length; i++) {            while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {                int tmp = nums[i];                nums[i] = nums[tmp - 1];                nums[tmp - 1] = tmp;            }        }        for (int i = 0; i < nums.length; i++) {            if (nums[i] != i + 1) {                list.add(i + 1);            }        }        return list;    }}