Leetcode 448. Find All Numbers Disappeared in an Array
Description Submission Solutions
- Total Accepted: 31266
- Total Submissions: 58997
- Difficulty: Easy
- Contributors: yuhaowang001
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]
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题解：
（1）， 不能申请额外的空间， 只能在原数组进行swap操作了。
（2）， 根据数据的类型，大小都是处于 [1, n], 进行位置操作。
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;
}
};