[LeetCode78]Remove Duplicates from Sorted Array II

数组去重算法

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本文介绍了一种算法，该算法可以在保持元素相对顺序的同时，将数组中重复出现超过两次的元素去除，并确保数组仍然保持有序状态。通过使用双指针技巧，此算法能够有效地处理这一问题。

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].

Analysis:

Similar with Remove Duplicates, here use to pointers to track duplicates numbers.

Here array is sorted, if the array is not sorted, we could use hashmap.

Java

public int removeDuplicates(int[] A) {
        if(A.length <3) return A.length;
		int pre =1, cur =1;
		int count =1;
		for(cur =1; cur< A.length ;cur++){
			if(A[cur] == A[cur-1]){
				if(count >=2){
					continue;
				}
				else{
					count++;
				}
			}else{
				count = 1;
			}
			A[pre] = A[cur];
			pre++;
		}
		return pre;
    }

c++

int removeDuplicates(int A[], int n) {
        if(n < 3) return n;
        int pre = 1, cur = 1;
        int count = 1;
        for(cur = 1; cur< n; cur++){
            if(A[cur] == A[cur-1]){
                if(count >=2){
                    continue;
                }else{
                    count++;
                }
            }else{
                count = 1;    
            }
            A[pre] = A[cur];
            pre++;
        }
        return pre;
    }