LEETCODE-- Find All Numbers Disappeared in an Array-优快云博客

本文介绍了一种在不额外占用空间且运行时间为O(n)的情况下找出数组中未出现的数字的方法。利用正负号标记法，通过将数组元素值作为下标进行操作，将已出现的数字对应位置标记为负数，从而高效地找到缺失的数字。

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]

这里使用到正负号标记法转载
大致思想：nums中元素值当作nums的下标进行操作，即凡是与nums中含有的元素值相同的下标下对应的元素变成负数（这里要注意如果前面的元素对后边的元素进行了改变，后边的元素再进行相应计算时有可能已经被变为负数，所以要在对元素进行计算时进行绝对值计算abs();）

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int len = nums.size();
        for(int i = 0; i < len; i++){
            cout << nums[i] << endl;
            int n = abs(nums[i]) - 1;
            nums[n] = -abs(nums[n]);
        }
        vector<int> dis;
        for(int j = 0; j < len; j++){
            if(nums[j] > 0)
                dis.push_back(j + 1);
        }   
        return dis;
    }
};