LEETCODE-- 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]

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
= -abs(nums
);
}
vector<int> dis;
for(int j = 0; j < len; j++){
if(nums[j] > 0)
dis.push_back(j + 1);
}
return dis;
}
};