Remove Duplicates from Sorted Array - LeetCode

https://leetcode.com/problems/remove-duplicates-from-sorted-array/?envType=study-plan-v2&envId=top-interview-150

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
• Return k.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

• 1 <= nums.length <= 3 * 104
• 100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

解析

这道题和27题类似，使用的方法也近似。很简单，直接写结果吧，还是two pointer的解法。

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        p1 = 0
        k = 0
        j = 100000

        for i in range(len(nums)):
            if nums[i] != j:
                j = nums[i]
                nums[p1] = nums[i]
                k += 1
                p1 += 1
            else:
                continue
        return k

上面的代码有优化的空间，首先k的不必要性在于 k = p1 + 1，其次，j也是不必须的，只要比较当前指针指的值和循环元素的值的异同就可以。

class Solution {
    public int removeDuplicates(int[] nums) {
        int p1 = 0;

        for (int i = 1; i < nums.length; i++) {
            if(nums[i]!=nums[p1]){
                p1++;
                nums[p1] = nums[i];
            }
        }
        return p1 + 1;
    }
}

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int p1 = 0;

        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i] != nums[p1]){
                p1++;
                nums[p1] = nums[i];
            }
        }
        return p1 + 1;
    }
};