[LeetCode] 26. Remove Duplicates from Sorted Array 有序数组中去除重复项-优快云博客

本文介绍了一种在原地(in-place)且仅使用O(1)额外空间的情况下，去除有序数组中重复元素的方法。通过使用双指针技巧，实现数组元素的遍历与更新，并提供Java、Python及C++等语言的实现示例。

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Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

去除有序数组中的重复项，返回新数组的长度，要求in-place，O(1)的额外空间。

解法：双指针Two Pointers，一个指针遍历数组所有元素，一个指针指向遍历后的不重复元素的边界。

Java:

class Solution {
    public static int removeDuplicatesNaive(int[] A) {
      if (A.length < 2)
        return A.length;

      int j = 0;
      int i = 1;

      while (i < A.length) {
        if (A[i] == A[j]) {
          i++;
        } else {
          j++;
          A[j] = A[i];
          i++;
        }
      }

      return j + 1;
    }
}

Python：

class Solution:
    # @param a list of integers
    # @return an integer
    def removeDuplicates(self, A):
        if not A:
            return 0
        
        last, i = 0, 1
        while i < len(A):
            if A[last] != A[i]:
                last += 1
                A[last] = A[i]
            i += 1
            
        return last + 1

C++:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int pre = 0, cur = 0, n = nums.size();
        while (cur < n) {
            if (nums[pre] == nums[cur]) ++cur;
            else nums[++pre] = nums[cur++];
        }
        return pre + 1;
    }
};

C++:

class Solution {
    public:
    int removeDuplicates(int A[], int n) {
        if(n < 2) return n;
        int id = 1;
        for(int i = 1; i < n; ++i) 
            if(A[i] != A[i-1]) A[id++] = A[i];
        return id;
    }
};

类似题目：

[LeetCode] 27. Remove Element 移除元素

[LeetCode] 283. Move Zeroes 移动零

[LeetCode] 80. Remove Duplicates from Sorted Array II 有序数组中去除重复项 II