[LeetCode] 26. Remove Duplicates from Sorted Array 有序数组中去除重复项

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 by modifying the input array in-place with O(1) extra memory.

Example:

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 new length.

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以元素，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;
    }
};

　　

类似题目：

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

　　

All LeetCode Questions List 题目汇总