[牛客网-Leetcode] #Array Medium remove-duplicates-from-sorted-array-ii

Remove duplicate numbers ii remove-duplicates-from-sorted-array-ii

Title description

Continue to think about the topic "Remove Duplicates":
Given a sorted array, remove duplicate elements.
What if the elements in the array are allowed to be repeated twice?
For example:
Given an ordered array A =[1,1,1,2,2,3],
the function you give should return length =5, and A becomes [1,1,2,2,3].

Follow up for “Remove Duplicates”:
What if duplicates are allowed at 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].

Problem-solving ideas

• Use fast and slow two pointers to record the non-duplicated area and traverse the array
• The pointer cnt is used to record the right subscript of the non-repeated area. The initial value is 0, which means there is one element.
• The pointer index is used to traverse the elements, the initial value is 1, and the traversal starts from the second element.
• maxnum is used to indicate the number of repetitions of the element, the initial value is 1, which means that the element only appears once
  • If A[cnt] == ​​A[index], then maxnum++, further judgment:
    • If maxnum<=2, then A[++cnt] = A[index],index++
    • If maxnum>2, skip, index++
  • If A[cnt]! = A[index], then A[++cnt] = A[index],index++, and because the elements are different, maxnum must be initialized to 1, which means that the number of repetitions of the next element is re-recorded

class Solution {
    
    
public:
    int removeDuplicates(int A[], int n) {
    
    
        if(n <= 1) return n;
        int cnt(0), maxnum(1);
        for(int index = 1; index < n; index ++) {
    
    
            if(A[cnt] == A[index]) {
    
    
                maxnum ++;
                if(maxnum <= 2) {
    
    
                    A[++ cnt] = A[index];
                }
            } else {
    
      //A[cnt] != A[index]
                maxnum = 1;
                A[++ cnt] = A[index];
            }
        }
        return cnt + 1;
    }
};