448. 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 a nd others appear once.

Find all the elements of [1, N] inclusive The does not appear in this array.

Could do it without extra space and in O (n) runtime? Assume the returned list does not count as extra space.

Example:

Input:

[4,3,2,7,8,2,3,1]

Output:

[5,6]

Idea:it ' s hard to solve without extra space and in O (n) without the condition 1≤a[i]≤ N.

Solution 1: for the number a[a[i]-1], if it was positive, assign it to its opposite number, else unchange it. The logic is if i+1 appear in the array, a[i] would be negative, so if a[i] are positive, i+1 doesn ' t appear in the array.

1 classSolution {2  Public:3vector<int> Finddisappearednumbers (vector<int>&nums) {4vector<int>Res;5          for(intI=0; I<nums.size (); i++){6             intIdx=abs (Nums[i])-1;7Nums[idx]= (nums[idx]>0)? (-Nums[idx]): Nums[idx];8         }9          for(intI=0; I<nums.size (); i++){Ten             if(nums[i]>0) Res.push_back (i+1); One         } A         returnRes; -     } -};

Solution 2:move Nums[i] from Positon I to their right position nums[i]-1, swap nums[i] and nums[nums[i]-1]. Finally, if I doesn ' t equal nums[i]-1, means i+1 doesn ' t appear in the array. Note the line 6-8

1 classSolution {2  Public:3vector<int> Finddisappearednumbers (vector<int>&nums) {4vector<int>Res;5          for(intI=0; I<nums.size (); i++){6             if(nums[i]!=nums[nums[i]-1]){7Swap (nums[i],nums[nums[i]-1]);8--i;9             }Ten         } One          for(intI=0; I<nums.size (); i++){ A             if(i!=nums[i]-1){ -Res.push_back (i+1); -             } the         } -         returnRes; -     } -};

Solution 3:add Nums[nums[i]-1] to N, use (nums[i]-1)%n to avoid the overflow of nums[i]-1. Finally, if nums[i]<=n, I+1 is the disappeard number.

1 classSolution {2  Public:3vector<int> Finddisappearednumbers (vector<int>&nums) {4vector<int>Res;5         intSize=nums.size ();6          for(intI=0; i<size;i++){7nums[(nums[i]-1)%size]+=size;//mod to avoid overflow8         }9          for(intI=0; i<size;i++){Ten             if(nums[i]<=size) { OneRes.push_back (i+1); A             } -         } -         returnRes; the     } -};

448. Find all Numbers disappeared in an Array