45 lines
1.4 KiB
C++
45 lines
1.4 KiB
C++
|
// Source : https://leetcode.com/problems/find-the-duplicate-number/
|
||
|
|
// Author : Calinescu Valentin
|
||
|
|
// Date : 2015-10-19
|
||
|
|
|
||
|
|
/***************************************************************************************
|
||
|
|
*
|
||
|
|
* Given an array nums containing n + 1 integers where each integer is between 1 and
|
||
|
|
* n (inclusive), prove that at least one duplicate number must exist.
|
||
|
|
* Assume that there is only one duplicate number, find the duplicate one.
|
||
|
|
*
|
||
|
|
* Note:
|
||
|
|
* You must not modify the array (assume the array is read only).
|
||
|
|
* You must use only constant, O(1) extra space.
|
||
|
|
* Your runtime complexity should be less than O(n2).
|
||
|
|
* There is only one duplicate number in the array, but it could be repeated more than
|
||
|
|
* once.
|
||
|
|
* Credits:
|
||
|
|
* Special thanks to @jianchao.li.fighter for adding this problem and creating all test
|
||
|
|
* cases.
|
||
|
|
*
|
||
|
|
***************************************************************************************/
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
/*
|
||
|
|
* Solutions
|
||
|
|
* =========
|
||
|
|
*
|
||
|
|
* A simple solution would be to sort the array and then look for equal consecutive elements.
|
||
|
|
*
|
||
|
|
* Time Complexity: O(N log N)
|
||
|
|
* Space Complexity: O(1)
|
||
|
|
*
|
||
|
|
*/
|
||
|
|
#include <algorithm>
|
||
|
|
class Solution {
|
||
|
|
public:
|
||
|
|
int findDuplicate(vector<int>& nums) {
|
||
|
|
sort(nums.begin(), nums.end());
|
||
|
|
for(vector<int>::iterator it = nums.begin(); it != nums.end(); ++it)
|
||
|
|
if(*it == *(it + 1))
|
||
|
|
return *it;
|
||
|
|
}
|
||
|
|
};
|