Leetcode ?448. Find All Numbers Disappeared in an Array
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Description?Submission?Solutions
- Total Accepted:?31266
- Total Submissions:?58997
- Difficulty:?Easy
- Contributors:?yuhaowang001
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Given an array of integers where 1 ≤ a[i] ≤?n?(n?= size of array), some elements appear twice and others appear once.
leetCode？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]
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an，題解：
（1）， 不能申請額外的空間， 只能在原數組進行swap操作了。?
（2）， 根據數據的類型，大小都是處于 [1, n], 進行位置操作。
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in、?
class Solution {
public:vector<int> findDisappearedNumbers(vector<int>& nums) {vector<int> ans;int i = 0; while(i<nums.size()){if(nums[i] != nums[ nums[i]-1 ]){swap(nums[i], nums[ nums[i]-1 ] ); }else{++i; }}for(int i=0; i<nums.size(); ++i){if(nums[i] != i+1){ans.push_back( i+1 ); }}return ans; }
};
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