leetcode--Find the Duplicate Number

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.

解题思路：先把数组排序，然后比较相邻的两个数，如果相等就代表是重复的。
c++:

#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;

int findDuplicate(vector<int>& nums)
{

    int n=nums.size();
    sort(nums.begin(),nums.end());

    for(int i=0; i<n; i++)
    {
        if(nums[i]==nums[i-1]) return nums[i];
    }

}
int main()
{
    vector<int> num;
    num.push_back(0);
    num.push_back(3);
    num.push_back(2);
    num.push_back(4);
    num.push_back(4);
    int s=findDuplicate(num);
    cout<<s<<endl;

}

上面的方法较慢，因为需要进行一次排序，时间复杂度较低的使用二分查找，如下：

#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;

int findDuplicate(vector<int>& nums)
{
    int n = nums.size(), i = 0, counter = 0;
    int left = 1, right = n - 1, mid = 0;

    while (left < right)
    {
        mid = left + (right - left) / 2;

        counter = 0;
        for (i = 0; i < n; ++i)
        {
            if (nums[i] <= mid)
                ++counter;
        }

        if (counter > mid)
            right = mid;
        else
            left = mid + 1;
    }

    return left;
}
int main()
{
    vector<int> num;
    num.push_back(0);
    num.push_back(2);
    num.push_back(2);
    num.push_back(4);
    num.push_back(6);
   int s= findDuplicate(num);
   cout<<s<<endl;


}