搜索元素的有效方法

You can do a linear search with steps that are often greater than 1. The crucial observation is that if e.g. array[i] == 4 and 7 hasn't yet appeared then the next candidate for 7 is at index i+3. Use a while loop which repeatedly goes directly to the next viable candidate.

Here is an implementation, slightly generalized. It finds the first occurrence of k in the array (subject to the +=1 restriction) or -1 if it doesn't occur:

#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;
}

output:

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

Your approach is too complicated. You don't need to examine every array element. The first value is 4, so 7 is at least 7-4 elements away, and you can skip those.

#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;
}

Program output:

Steps 4, index 11

Edit: improved after comments from @Martin Zabel.

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.

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.)

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.

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;
}

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;


}

const findMeAnElementsFunkyArray = (arr, ele, i) => {
const elementAtCurrentIndex = arr[i];


const differenceBetweenEleAndEleAtIndex = Math.abs(
ele - elementAtCurrentIndex
);


const hop = i + differenceBetweenEleAndEleAtIndex;


if (i >= arr.length) {
return;
}
if (arr[i] === ele) {
return i;
}


const result = findMeAnElementsFunkyArray(arr, ele, hop);


return result;
};


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


const answer = findMeAnElementsFunkyArray(array, 7, 0);


console.log(answer);

Wanted to include a recursive solution to the problem. Enjoy