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]

查找消失的元素，重点在于： 1=<a[i] <=n , O(n)的时间复杂度，最先想到的是直接在数组上标记为负，因为之前做 

Find All Duplicates in an Array(查找数组中出现两次的数据)时，就是这种解法。所以一上来直接模拟下AC了。

/**
* @param {number[]} nums
* @return {number[]}
*/
var findDisappearedNumbers = function(nums) {
var arr = [];
for(var i =0 ;i<nums.length ;i++){
var num = Math.abs(nums[i]);
if(nums[num-1]>0){
nums[num-1] *= -1;
}
}
for(var i =0 ;i<nums.length ;i++){
if(nums[i] > 0){
arr.push(i+1);
}
}
return arr;
};解法二：（网上看到的）将nums[i]置换到其对应的位置nums[nums[i]-1]上去，比如对于没有缺失项的正确的顺序应该是[1, 2, 3, 4, 5, 6, 7, 8]，而我们现在却是[4,3,2,7,8,2,3,1]，我们需要把数字移动到正确的位置上去，比如第一个4就应该和7先交换个位置，以此类推，最后得到的顺序应该是[1, 2, 3, 4, 3, 2, 7, 8]，我们最后在对应位置检验，如果nums[i]和i+1不等，那么我们将i+1存入结果res中即可【转】

var arr = [];
for(var i =0 ;i<nums.length ; i++){
var num = nums[i]-1;
if(nums[i]!= nums[num]){
var tt = nums[num];
nums[num] = nums[i];
nums[i] = tt;
--i;
console.log(nums);
}
}
for(var i =0 ;i<nums.length ; i++){
if(nums[i] != i+1){
arr.push(i+1);
}
}
return arr;