Offer surface prove safety questions in duplicate numbers 3_2 array (not modify the original contents of the array)

Problem: In a case where editing is not given array , an array of length n + 1 is found in any one of the duplicate numbers, all numbers within the range of 1 ~ n.

Length of the array, the array: input

Output: duplicate numbers

Ideas: Unable to modify the original array, face questions 3_1 algorithm ideas can not be used. The conditions of this question, must be a duplicate number, if an open array of size n, the original digital array into a new array table for position values, flower O (n) time complexity can find duplicate numbers. However, the space complexity is also O (n).

This article sacrifice time complexity for space complexity, similar dichotomy algorithm to find ideas are as follows:

Using the target properties of the array (index based on the elements corresponding positioning).

The numbers 1 ~ n (of length n + 1) is divided into two from the middle portion of the number m, the front half 1 ~ m, the back half of the m + 1 ~ n. If the number of digits exceeds 1 ~ m m, then it must contain half of the interval in duplicate numbers. In this case, the interval comprising repeating continues into two digits, until a duplicate numbers.

Code:

#include <iostream>
using namespace std;

int countRange(const int* numbers, int length, int start, int end);

int getDuplication(const int* numbers, int length)
{
	if(numbers==nullptr || length<=0)
		return -1;
	int start = 1;
	int end = length-1;
	while(end >= start)
	{
		int middle = ((end - start)>>1)+start;
		int count = countRange(numbers, length, start, middle);
		if(end==start)
		{
			if(count> 1)
				return start;
			else
				break;
		}
		if(count>(middle-start+1))
			end =middle;
		else
			start=middle+1;
	}
	return -1;
}

int countRange(const int* numbers, int length, int start, int end)
{
	if(numbers==nullptr)
		return 0;
	int count = 0;
	for(int i=0; i<length; i++)
	{
		if(numbers[i]>=start&&numbers[i]<=end)
			count+=1;
	}
	return count;
}
int main(int argc, char const *argv[])
{
	int numbers[] = {0,3,2,1,3};
	int result;
	result = getDuplication(numbers,5);
	if(result!=-1)
		cout<<result<<endl;
	else
		cout<<"not found"<<endl;
	return 0;
}

Complexity Analysis: time complexity of O (nlogn), the spatial complexity is O (1).

Expansion: the algorithm can not find all duplicate numbers.

For example, the array {2,3,5,7,4,3,2,6}, the numerical range of 1 to 2 occurs twice, the algorithm can not be determined once each occurrence 1 and 2, a digital still appeared 2 times.