leetcode-448. Find All Numbers Disappeared in an Array-优快云博客

本文介绍了一种在O(n)时间内查找指定范围内未出现在数组中的所有整数的方法，并提供了两种实现方案，一种使用额外数组标记已出现元素，另一种通过修改原数组达到空间优化。

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448. Find All Numbers Disappeared in an Array

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]

第一种解法是我自己一开始写的，很一般，solution里的解法不错。

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        int[] tep = new int[nums.length];
        for(int i = 0;i<nums.length;i++){
            tep[nums[i]-1]=1;
        }
        List<Integer> list2 = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if(tep[i]==0) list2.add(i+1);
        }
        return list2;
    }
}

Runtime: 14 ms

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;
    }

这种的空间复杂度要小一些