leetcode: 448. Find All Numbers Disappeared in an Array

最新推荐文章于 2025-12-18 17:00:31 发布

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#leetcode

本文介绍了一个LeetCode上的经典算法题，即如何找出数组中消失的数字。通过巧妙地利用数组自身作为标记，实现O(1)空间复杂度和O(n)时间复杂度的要求。文章详细解释了解决方案，并附带了C++实现代码。

题目解析：

题目链接：https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/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]

题目有很多解法，难的是O(1)空间复杂度和O（n）时间复杂度。用到题目给出的条件，将出现过的数值作为数组的位置标志，将对应位置置为0，最后数组中不为0的位置就是没出现过的数字。

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        int i = 0;
        while(i<nums.size())
        {
            if(nums[i] != 0&&nums[nums[i]-1]>0)
            {
                int a = nums[i]-1;
                swap(nums[nums[i]-1],nums[i]);
                nums[a] = 0;
            }
            else
            {
                i++;
            }
        }
        for(int j = 0; j < nums.size(); j++)
        {
            if(nums[j]>0)
                res.push_back(j+1);
        }
        return res;
    }
};