leetcode--Find the Duplicate Number

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本文介绍了一种在不修改原始数组且仅使用常数级额外空间的情况下找出数组中重复数字的方法。提出了两种解决方案，一种通过排序后比较相邻元素实现，另一种利用二分查找提高效率。

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;


}