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:
1 class Solution { 2 public: 3 vector<int> findDisappearedNumbers(vector<int>& a) 4 { 5 vector<int> vecDisappear;vecDisappear.clear(); 6 int n = a.size(); 7 for(int i = 0;i<n;) 8 { 9 if(a[i] != a[a[i]-1]) 10 { 11 swap(a[i],a[a[i]-1]); 12 } 13 else 14 i++; 15 } 16 for(int i = 0;i<n;i++) 17 { 18 if(a[i]!=i+1) 19 { 20 vecDisappear.push_back(i+1); 21 } 22 } 23 return vecDisappear; 24 }
思路：遍历vector，对于i上的数字而言，确保其上出现的数字a[i]已经出现在正确的位置a[a[i]-1]上，否则交换二者位置。
solution2:
class Solution { public: vector<int> findDisappearedNumbers(vector<int>& a) { vector<int> vecDisappear;vecDisappear.clear(); for(int i = 0;i<a.size();i++) { auto index = abs(a[i])-1; if(a[index]>0) { a[index]=-a[index]; } } for(int i = 0;i<a.size();i++) { if(a[i]>0) vecDisappear.push_back(i+1); } return vecDisappear; } };
思路：对于i上出现的数字a[i]，即表示索引a[i]-1位置上的数字已成功出现。