Java code implementation deletes a repeating element inside an ordered array __java

Relaxed so many days, finally put the blog back to pick up, after the day to maintain 3 more, refueling refueling.

The algorithm for this implementation is to delete the repeating elements in an ordered array

Train of thought: An array is ordered, so the algorithm is relatively simple to implement, because only compare the array adjacent to two digits, there are two situations

1: If there is no element in the array or there is only one element, then there is no need to compare, directly return the length of the array;

2: If the length of the array is greater than one, then you need to compare the adjacent two elements of the array. If it is equal then the pointer of the latter element moves back one bit, and then the pointer of the previous element is then moved backward, assigning the number indicated by the current second element pointer to the position indicated by the previous element pointer. Then the latter element pointer continues to add one. If the adjacent elements are unequal, the immediate preceding element pointer is superimposed with the latter element, and then the latter element pointer continues to add one.

The code is as follows:

public class Solution {
public int removeduplicates (int[] A) {
int i=0;
int j=1;
if (a.length==0| | A.length==1) {
return a.length;
}
while (I<a.length&&j<a.length) {
if (A[i]==a[j]) {
j + +;
}else{
A[++i]=a[j++];
}

}
return i+1;
}
}