448. Find All Numbers Disappeared in an Array

QUESTION

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]

THOUGHT I

用一个1-n的数组来代替标志位，开始的时候都赋负值，然后把原数组的数放在应该的位置，然后再重新遍历标志数组，如果为负数则表示缺失。

CODE

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

RESULT

time complexity is O(n),space complexity is O(n);

THOUGHT II

同样是用标志位的方法去做，把原数组中出现的数其应该所在的位置上置为负值，然后重新遍历如果大于0，则表示从未出现过。

CODE

public class Solution {    public List<Integer> findDisappearedNumbers(int[] nums) {        List<Integer> res = new ArrayList<Integer>();        if(nums == null || nums.length == 0)            return res;        int n = nums.length;        for(int i = 0;i < n;i++){            int val = Math.abs(nums[i]) - 1;            if(nums[val] > 0)                nums[val] = -nums[val];        }        for(int i = 0;i < n;i++){            if(nums[i] > 0)                res.add(i + 1);        }        return res;    }}

RESULT

time complexity is O(n),space complexity is O(1);