Interview Question 3 (b): Do not modify the array to find duplicate numbers

Interview Question 3 (b): Do not modify the array to find duplicate numbers
Title: All numbers in an array of length n+1 are within the range of 1 to N, so the array
A few numbers are duplicated. Please find any duplicate numbers in the array, but you cannot modify the input
Array. For example, if you enter an array of length 8 {2, 3, 5, 4, 3, 2, 6, 7}, then the corresponding
The output is a repeating number 2 or 3.

Problem Solving Ideas:

You cannot modify an array, you can create a secondary array of length n+1, and the space complexity is O (n).

If you use time to change space, you can use the idea of two-point search.

The element range is 1~n, but there are n+1 elements that indicate that a number has been duplicated.

The median number m is divided into two parts, 1~m and M+1~n.

If the number of elements in the 1~m exceeds m, the repeating element is divided in the first half, again 1~m.

Repeat this process until you finally find the duplicate element.

Pseudo code:

if(Invalid input parameter)return-1;
intstart=1;intend=length-1; while(start<=end) {    intmiddle=interval midpoint; intCount=the number of elements between start and middle; if(the upper and lower bounds of the interval are equal) {if(count greater than 1)returnstart; Else            return-1; }    if(Count> (middle-start+1)) End=Middle; ElseStart=middle+1;}return-1;

C + + code:

intGetduplication (Const int* Numbers,intlength) {    //Verifying the validity of input parameters    if(numbers==nullptr| | length<0){        return-1; }     for(intI=0; i<length;i++){        if(numbers[i]<1|| numbers[i]>length-1){            return-1; }    }    //finding duplicate elements in two points    intstart=1; intend=length-1;  while(end>=start) {        intMiddle= (End-start) >>1)+start; intCount=Countrange (Numbers,length,start,middle); //the upper and lower bounds of the interval are equal        if(end==start) {            //the number of elements is greater than 1, successful lookups            if(count>1)                returnstart; Else                 Break; }        //range Upper and lower bounds, continuation of binary search        if(Count> (middle-start+1)) End=Middle; ElseStart=middle+1; }    return-1;}//get the number of interval elementsintCountrange (Const int* Numbers,intLengthintStartintmiddle) {    if(numbers==nullptr| | length<0){        return-1; }        intCount=0;  for(intI=0; i<length;i++){        if(numbers[i]>=start&&numbers[i]<=middle) Count++; }    returncount;}

Resources:

Sword point 3 (b)

Interview Question 3 (b): Do not modify the array to find duplicate numbers