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]
方法一：用hash表标记。
1 class Solution { 2 public: 3 vector<int> findDisappearedNumbers(vector<int>& nums) { 4 vector<int> result; 5 vector<bool> hash(nums.size() + 1, false); 6 for(int i = 0; i < nums.size(); i++){ 7 hash[nums[i]] = true; 8 } 9 for(int i = 1; i < hash.size(); i++){ 10 if(hash[i] == false){ 11 result.push_back(i); 12 } 13 } 14 return result; 15 } 16 };
方法二：不用hash表。思路就是：比如有4，那么将第4个元素变为负数（不能变为0），最后再遍历一次数组，哪个位置上的数为正，则那个位置对应的数不存在。
1 class Solution { 2 public: 3 vector<int> findDisappearedNumbers(vector<int>& nums) { 4 vector<int> result; 5 for(int i = 0; i < nums.size(); i++){ 6 if(nums[abs(nums[i]) - 1] > 0) 7 nums[abs(nums[i]) - 1] *= -1; 8 } 9 for(int i = 0; i < nums.size(); i++){ 10 if(nums[i] > 0){ 11 result.push_back(i + 1); 12 } 13 } 14 return result; 15 } 16 };
448. Find All Numbers Disappeared in an Array
原文：http://www.cnblogs.com/qinduanyinghua/p/6350162.html