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

最新推荐文章于 2025-12-18 16:28:22 发布

最新推荐文章于 2025-12-18 16:28:22 发布 · 79 阅读

文章标签：

#c/c++

本文介绍两种使用C++去除已排序数组中重复元素的方法，并确保每个元素仅出现一次。方法一利用双指针技巧，方法二通过检测重复并调整数组长度。此外，还展示了如何使用C++ STL中的unique函数实现相同功能。

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;
	}
};