요소를 효율적으로 검색하는 방법

요소를 효율적으로 검색하는 방법

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최근에 인터뷰를했는데 그들은 " 검색 "질문을했습니다.
질문은 :

각 요소가 인접한 요소 중 하나 +1이거나 -1비교 되는 (양수) 정수 배열이 있다고 가정 합니다.

예:

array = [4,5,6,5,4,3,2,3,4,5,6,7,8];

이제 7위치를 검색 하고 반환합니다.

나는이 대답을했다 :

값을 임시 배열에 저장하고 정렬 한 다음 이진 검색을 적용합니다.

요소가 발견되면 임시 배열에서 해당 위치를 반환합니다.
(번호가 두 번 발생하면 첫 번째 발생을 반환)

그러나 그들은이 답변에 만족하지 않는 것 같습니다.

정답은 무엇입니까?

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1보다 큰 단계로 선형 검색을 수행 할 수 있습니다. 중요한 관찰은 eg array[i] == 4와 7이 아직 나타나지 않은 경우 7에 대한 다음 후보가 index에 있다는 것 i+3입니다. 다음 실행 가능한 후보로 반복적으로 직접 이동하는 while 루프를 사용하십시오.

다음은 약간 일반화 된 구현입니다. k배열에서 첫 번째 발생 (+ = 1 제한에 따라) 또는 -1발생하지 않는 경우를 찾습니다 .

#include <stdio.h>
#include <stdlib.h>

int first_occurence(int k, int array[], int n);

int main(void){
    int a[] = {4,3,2,3,2,3,4,5,4,5,6,7,8,7,8};
    printf("7 first occurs at index %d\n",first_occurence(7,a,15));
    printf("but 9 first \"occurs\" at index %d\n",first_occurence(9,a,15));
    return 0;
}

int first_occurence(int k, int array[], int n){
    int i = 0;
    while(i < n){
        if(array[i] == k) return i;
        i += abs(k-array[i]);
    }
    return -1;
}

산출:

7 first occurs at index 11
but 9 first "occurs" at index -1

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접근 방식이 너무 복잡합니다. 모든 배열 요소를 검사 할 필요는 없습니다. 첫 번째 값은 4그래서 7입니다 적어도 7-4 멀리 요소, 당신은 사람들을 건너 뛸 수 있습니다.

#include <stdio.h>
#include <stdlib.h>

int main (void)
{
    int array[] = {4,5,6,5,4,3,2,3,4,5,6,7,8};
    int len = sizeof array / sizeof array[0];
    int i = 0;
    int steps = 0;
    while (i < len && array[i] != 7) {
        i += abs(7 - array[i]);
        steps++;
    }

    printf("Steps %d, index %d\n", steps, i);
    return 0;
}

프로그램 출력 :

Steps 4, index 11

편집 : @Raphael Miedl 및 @Martin Zabel의 의견 이후 개선되었습니다.

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A variation of the conventional linear search could be a good way to go. Let us pick an element say array[i] = 2. Now, array[i + 1] will either be 1 or 3 (odd), array[i + 2] will be (positive integers only) 2 or 4 (even number).

On continuing like this, a pattern is observable - array[i + 2*n] will hold even numbers and so all these indices can be ignored.

Also, we can see that

array[i + 3] = 1 or 3 or 5
array[i + 5] = 1 or 3 or 5 or 7

so, index i + 5 should be checked next and a while loop can be used to determine the next index to check, depending on the value found at index i + 5.

While, this has complexity O(n) (linear time in terms of asymptotic complexity), it is better than a normal linear search in practical terms as all the indices are not visited.

Obviously, all this will be reversed if array[i] (our starting point) was odd.

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The approach presented by John Coleman is what the interviewer was hoping for, in all probability.
If you are willing to go quite a bit more complicated, you can increase expected skip length:
Call the target value k. Start with the first element's value v at position p and call the difference k-v dv with absolute value av. To speed negative searches, have a peek at the last element as the other value u at position o: if dv×du is negative, k is present (if any occurrence of k is acceptable, you may narrow down the index range here the way binary search does). If av+au is greater than the length of the array, k is absent. (If dv×du is zero, v or u equals k.)
Omitting index validity: Probe the ("next") position where the sequence might return to v with k in the middle: o = p + 2*av.
If dv×du is negative, find k (recursively?) from p+av to o-au;
if it is zero, u equals k at o.
If du equals dv and the value in the middle isn't k, or au exceeds av,
or you fail to find k from p+av to o-au,
let p=o; dv=du; av=au; and keep probing.
(For a full flash-back to '60ies texts, view with Courier. My "1st 2nd thought" was to use o = p + 2*av - 1, which precludes du equals dv.)

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STEP 1

Start with the first element and check if it's 7. Let's say c is the index of the current position. So, initially, c = 0.

STEP 2

If it is 7, you found the index. It's c. If you've reached the end of the array, break out.

STEP 3

If it's not, then 7 must be atleast |array[c]-7| positions away because you can only add a unit per index. Therefore, Add |array[c]-7| to your current index, c, and go to STEP 2 again to check.

In the worst case, when there are alternate 1 and -1s, the time complexity may reach O(n), but average cases would be delivered quickly.

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Here I am giving the implementation in java...

public static void main(String[] args) 
{       
    int arr[]={4,5,6,5,4,3,2,3,4,5,6,7,8};
    int pos=searchArray(arr,7);

    if(pos==-1)
        System.out.println("not found");
    else
        System.out.println("position="+pos);            
}

public static int searchArray(int[] array,int value)
{
    int i=0;
    int strtValue=0;
    int pos=-1;

    while(i<array.length)
    {
        strtValue=array[i];

        if(strtValue<value)
        {
            i+=value-strtValue;
        }
        else if (strtValue==value)
        {
            pos=i;
            break;
        }
        else
        {
            i=i+(strtValue-value);
        }       
    }

    return pos;
}

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Here is a divide-and-conquer style solution. At the expense of (much) more bookkeeping, we can skip more elements; rather than scanning left-to-right, test in the middle and skip in both directions.

#include <stdio.h>                                                               
#include <math.h>                                                                

int could_contain(int k, int left, int right, int width);                        
int find(int k, int array[], int lower, int upper);   

int main(void){                                                                  
    int a[] = {4,3,2,3,2,3,4,5,4,5,6,7,8,7,8};                                   
    printf("7 first occurs at index %d\n",find(7,a,0,14));                       
    printf("but 9 first \"occurs\" at index %d\n",find(9,a,0,14));               
    return 0;                                                                    
}                                                                                

int could_contain(int k, int left, int right, int width){                        
  return (width >= 0) &&                                                         
         (left <= k && k <= right) ||                                            
         (right <= k && k <= left) ||                                            
         (abs(k - left) + abs(k - right) < width);                               
}                                                                                

int find(int k, int array[], int lower, int upper){                              
  //printf("%d\t%d\n", lower, upper);                                            

  if( !could_contain(k, array[lower], array[upper], upper - lower )) return -1;  

  int mid = (upper + lower) / 2;                                                 

  if(array[mid] == k) return mid;                                                

  lower = find(k, array, lower + abs(k - array[lower]), mid - abs(k - array[mid]));
  if(lower >= 0 ) return lower;                                                    

  upper = find(k, array, mid + abs(k - array[mid]), upper - abs(k - array[upper]));
  if(upper >= 0 ) return upper;                                                  

  return -1;                                                                     

}

참고URL : https://stackoverflow.com/questions/34481582/efficient-way-to-search-an-element