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448. Find All Numbers Disappeare

448. Find All Numbers Disappeare

作者: namelessEcho | 来源:发表于2017-09-14 16:45 被阅读0次

    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.

    应该还是基准排序,主要问题在于如何判断出现两次,这里在第二次访问时可以发现索引上的数被置负了,所以可以判断出现两次。

    class Solution {
        public List<Integer> findDisappearedNumbers(int[] nums) {
            List<Integer> result = new ArrayList<>();
            for(int i = 0 ;i<nums.length;i++)
            {
                int index = nums[i]<0?-nums[i]-1:nums[i]-1;
                if(nums[index]>0)
                    nums[index]=-nums[index];
            }
             for(int i = 0 ;i<nums.length;i++)
             {
                 if(nums[i]>0)
                     result.add(i+1);
             }
            return result ;
        }
    }
    

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