[leet code] Remove Duplicates from Sorted Array I & II

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最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： interview preparation 文章标签： algorithm array leetcode java

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本文详细介绍了如何通过双指针法解决数组去重问题，包括基本版和允许最多两次重复的进阶版。通过实例演示，帮助理解如何在不额外分配空间的情况下修改原始数组。

Remove Duplicates from Sorted Array I :

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array A = [1,1,2],

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

Remove Duplicates from Sorted Array II:

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:

Key points of of these 2 problems: use 2 pointers to overwrite the original array according to the problem requirement. One pointer is for the original array, +1 every time. Another pointer is for the overwritten array, +1 when current number not duplicated or not duplicated 2 times (problem II).

Remove Duplicates from Sorted Array I :

public class Solution {
    public int removeDuplicates(int[] A) {
        // exception case
        if (A.length <= 1) return A.length;

        int oPointer=1; // pointer for original array
        int nPointer=0; // pointer for overwritten array
        
        while (oPointer<A.length){
            if(A[oPointer] == A[nPointer]) oPointer++; // duplicate case
            else{
                nPointer++;
                A[nPointer] = A[oPointer];
                oPointer++;
            }
        }
        return nPointer+1;
    }
}

Remove Duplicates from Sorted Array II:

public class Solution {
public int removeDuplicates(int[] A) {
        if (A.length<3) return A.length;
        
        int nPointer = 1; // pointer for overwritten array
        int oPointer = 2; // pointer for original array
        while(oPointer<A.length){
            if(A[oPointer]==A[nPointer] && A[oPointer]==A[nPointer-1]) 
                oPointer++;
            else{
                nPointer++;
                A[nPointer] = A[oPointer];
                oPointer++;
            }
        }
        return nPointer+1;
    }
}