[Baidu] Array A of any two adjacent element size difference 1, in which to find a number

I. Source and description of the problem

Today saw July's microblog, found the July problem, there is this problem, very interesting.

The size of any two adjacent elements in array a differs by 1, and is given the array A and target integer t to find the position of T in array A. Arrays: [1,2,3,4,3,4,5,6,5], locate 4 in the array.

two. Algorithm analysis and implementation

the worst time complexity of the topic is also O (N) (increment or decrement), so the focus is on finding a way to reduce the number of comparisons as much as possible. Arrays : [1,2,3,4,3,4,5,6,5], locate 4 in the array. 4 and 1 Compare, the difference is 3, so even if the best case (increment or decrement), 4 is in a[3] position, you can skip a[1]a[2]. This may save time if a particular array (target value and a[1) is quite different.

So the rule: for the target T, starting from the current position a[i] comparison, the next possible position is I = ABS (A[I]-T).

public class Solution {public    static vector<integer> ve = new vector<integer> ();    public static void Find (int num[], int n, int target) {        if (n <= 0) return;        for (int i = 0; i < n;) {            if (num[i] = = target) {                ve.add (i);                i + = 2;            }            else i + = Math.Abs (Num[i]-target);        }        return;    }    public static void Main (String args[]) {        ve.clear ();        int num[] = {1, 2, 3, 2, 3, 4, 3, 2, 3};        int target = 4;        Find (num, num.length, target);        for (int i:ve)            System.out.println (i + "");}    }

Why +2? such as "4,3,4". +1 certainly not.

[Baidu] Array A of any two adjacent element size difference 1, in which to find a number