leetcode-448-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.

Example:

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Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]
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class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> ret = new ArrayList<Integer>();
        
        for(int i=0; i<nums.length; i++)
        {
            int val = Math.abs(nums[i]) - 1;
            if(nums[val] > 0)
                    nums[val] = -nums[val];
        }
        
        for(int i=0; i<nums.length; i++)
        {
            if(nums[i] > 0)
                ret.add(i+1);
        }
        
        return ret;
    }
}
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Take input [4.3.2.7.8.2.3.1] as an example, by subtracting 1 it becomes [3.2.1.6.7.1.2.0] which is an array of index.
For the first iteration
when i = 0 , it marks the nums[3] as negative, the array become [4.3.2.-7.8.2.3.1].
when i = 1, it marks the nums[2] as negative, the array become [4.3.-2.-7.8.2.3.1].
when i = 2, it marks the nums[1] as negative, the array become [4.-3.-2.-7.8.2.3.1].
when i = 3, it marks the nums[6] as negative, the array become [4.-3.-2.-7.8.2.-3.1].
...
...
when i = 6, it marks the nums[0] as negative, the array become [-4.-3.-2.-7.8.2.-3.-1].

For the second iteration
find nums[4] = 8 and nums[5] = 2 which > 0;
which means 4 and 5 are not in the index array[3.2.1.6.7.1.2.0].
by adding 1, 5 and 6 are not in the input[4.3.2.7.8.2.3.1]
return[5.6]
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