leetCode-Find All Numbers Disappeared in an Array

Posted 2020-10-15 kevincong

Description:
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]

My Solution:

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        int len1 = nums.length;
        Set<Integer> set = new HashSet<Integer>();
        Set<Integer> set1 = new HashSet<Integer>();
        for(int i = 0;i < len1;i++){
            set.add(Integer.valueOf(i + 1));
            set1.add(Integer.valueOf(nums[i]));
        }
        set.removeAll(set1);
        return new ArrayList<Integer>(set);
    }
}

Another Better Solution:

public class Solution {

    public List<Integer> findDisappearedNumbers(int[] nums) {

        List<Integer> list = new ArrayList<>();

        boolean[] a = new boolean[nums.length + 1];

        for(int i = 0; i < nums.length; i++) {
            a[nums[i]] = true;
        }

        for(int i = 1; i < a.length; i++) {
            if(!a[i]){
                list.add(i);
            }
        }

        return list;
    }
}

 

总结：建立一个boolean数组,对应下标为true表示数值存在,false表示不存在,遍历nums,将nums值作为下标，赋为true。遍历boolean数组,为false值的下标即为不存在，添加到ArrayList即可(注意,boolean数组的长度为nums的长度加一，方便一一对应)。
重要的是要观察到nums值与即为boolean数组下标。