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.

此方法修改了数组，因此作为一种实现可能。 

int compare (const void * a, const void * b)
{
    if(*(int *)a < *(int *)b)
        return -1;
    else if(*(int *)a > *(int *)b)
        return 1;
    else 
        return 0;
}

int findDuplicate(int* nums, int numsSize) 
{
    int i = 1;
    qsort (nums, numsSize, sizeof(int), compare);
    for(i = 1;i < numsSize;i++)
    {
        if(nums[i] == nums[i-1])
            break;
    }
    
    return nums[i];
    
}

一种比较巧妙的实现方法： 

int findDuplicate(int* nums, int numsSize) 
{
    int low=1;
    int high=numsSize - 1;
    int mid;
    
    while(low<high)
    {
        mid=(low+high)/2;
        int count=0;
        for(int i = 0;i < numsSize;i++)
        {
            if(nums[i] <= mid) count++;
        }
        
        if(count>mid) 
            high=mid;
        else 
            low=mid+1; 
    }
    
    return low;
}