LeetCode Remove Duplicates from Sorted Array 移除有序数列的重复元素

最新推荐文章于 2025-11-28 16:17:07 发布

最新推荐文章于 2025-11-28 16:17:07 发布 · 106 阅读

文章标签：

#c/c++

本文介绍了一种方法来去除给定数组中的重复元素，并返回处理后的数组长度，同时保持空间复杂度为常数。

Given a sorted array, remove the duplicates in place such that each element appear onlyonceand 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].

小心令人抓狂的下标！

方法一：

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n<=1) return n;
        int j = 0;
        for (int i = 1; i < n; ++i) {
            if (A[i] != A[j]) {
                A[++j] = A[i];
            }
        }
        return ++j;
    }
};

方法二：

class Solution {
public:
	int removeDuplicates(int A[], int n) {
		if(n<=1) return n;
		int dup = 0;
		int i = 1;
		for (; i < n; i++)
		{
			if(A[i] == A[i-1]) dup++;
			else if(dup != 0)
			{
				for (int j = i;  j < n; j++)
				{
					A[j-dup] = A[j];
				}
				n -= dup;
				i -= dup;
				dup = 0;
			}
		}
		if(dup != 0)
		{
			A[n-dup-1] = A[n-1];
			n -= dup;
			dup = 0;
		}
		return n;
	}
};

C++ STL

template <class ForwardIterator>
  ForwardIterator unique (ForwardIterator first, ForwardIterator last)
{
  if (first==last) return last;

  ForwardIterator result = first;
  while (++first != last)
  {
    if (!(*result == *first))  // or: if (!pred(*result,*first)) for version (2)
      *(++result)=*first;
  }
  return ++result;
}

//2014-1-25
class Solution {
public:
	int removeDuplicates(int A[], int n) 
	{
		int i = 1, j = 1;
		for ( ; j < n; j++)
		{
			if (A[j] != A[j-1]) A[i++] = A[j];
		}
		return n? i:0;
	}
};