[leetcode-448]Find All Numbers Disappeared in an Array

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本文介绍了一种在O(n)运行时间内且不使用额外空间寻找数组中消失元素的方法。该算法巧妙地利用数组元素作为索引来标记已出现的数字，通过反转对应位置的数值来定位未出现的元素。

Problem 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]

Solution

For an array whoes elements are in range of 1 to n, considerate to use the element as the index to identify the missed or duplicated elements.
The key point is locate and mark, with the following logic:

array[array[i] - 1] *= -1; // where i = 0, 1, 2, 3... , n-1

code:

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