LeetCode - 448. Find All Numbers Disappeared in an Array

最新推荐文章于 2022-06-14 17:41:04 发布

原创最新推荐文章于 2022-06-14 17:41:04 发布 · 306 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种高效查找数组中缺失元素的方法。对于长度为n的数组，其中包含1到n的整数，某些数出现两次而其他数仅出现一次。通过使用哈希表，我们可以在O(n)时间内找到所有未出现在数组中的数。

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]

建立一个大小为size + 1的哈希表，O（n）的时间填充哈希表，O（n）的时间扫描哈希表并添加元素。

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> vec;
        int size = nums.size();
        int* table = new int[size + 1] { 0 };
        
        for( int i = 0; i < size; i++ ) {
            table[nums[i]]++;
        }
        
        for( int i = 1; i <= size; i++ ) {
            if( table[i] == 0 ) vec.push_back( i );
        }
        
        return vec;
    }
};