(Daily algorithm) Leetcode---Remove duplicates from Sorted Array II (remove duplicate element II)

Remove duplicates from Sorted Array II

Leetcode

Topic:

Follow up for "Remove duplicates":
What if duplicates is allowed at the most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

Given an ordered table, each element appears up to two times. Remove the extra elements and return the number of remaining elements.

Ideas:

Two pointers, one pointing to the next position in the legal position （index） , and the other for traversing the element (i) . The invalid element is between two pointers.

This is the time i to determine whether the element pointed to is the same as index - 2 the element pointed to.

• Same: i the description points to an element that is superfluous and invalid. imove back.
• Not the same: indicates that i the element pointed to is valid, that is, not more than 2 occurrences. At this point, i replace it index . indexneed to move back

can be viewed in the markdown documentation: Click to open the link code as follows:

  
 

   
  

  
class Solution {
   
  

  
public:
   
  

  
 int removeDuplicates(int A[], int n) {
   
  

  
 int deleted = 0; //已经删除的元素个数
   
  

  
 if(n < 3) //最多含有2个元素，一定满足题意
   
  

  
 return n;
   
  

  
 int index = 2; //可能不合法的位置
   
  

  
 for(int i = 2; i < n; i++)
   
  

  
 {
   
  

  
 if(A[i] != A[index - 2] )
   
  

  
 A[index++] = A[i];
   
  

  
 }
   
  

  
 return index;
   
  

  
 }
   
  

  
};
  
 

(Daily algorithm) Leetcode---Remove duplicates from Sorted Array II (remove duplicate element II)