Title
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]
题目大意：找失去的数字，两个
Solution
交换到相应位置
class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& nums) {
vector<int> res;
for (int i = 0; i < nums.size(); i++) {
while (nums[i]!=i+1&&nums[nums[i]-1]!=nums[i])
{
swap(nums[i], nums[nums[i] - 1]);
}
}
for (int i = 0; i < nums.size(); i++) {
if (nums[i] != i + 1) {
res.push_back(i+1);
cout << i+1 << " ";
}
}
return res;
}
};