[LeetCode Road] Remove Duplicates from Sorted Array - Solution/C++

26. Remove Duplicates from Sorted Array

Question:

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns 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.
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 a 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]);
}

Example:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: 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.

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 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.

Constraints:

0 <= nums.length <= 3 * 10^4

-10^4 <= nums[i] <= 10^4

nums is sorted in ascending order.

Source code

Version 1

Idea:
This is a simple solution. Remain the first element, so a count index is start from 1. Use a loop to compare with each elements, if it find the result is true, store the element and shift a count index.

Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (!nums.size()) return 0;
        int count = 1;
        for(int i = 1; i < nums.size(); i++){
            if(nums[i - 1] != nums[i])
                nums[count++] = nums[i];
        }
        return count;
    }
};

Version 2

Idea:
This better solution is refered to Huahua's Tech Road.
At first, nums[count++] = nums[i] aims to store the first element and shift a count index. Declare a new variable j as i + 1 to find the element which is not equal to nums[i]. Then, if it is found, use i = j to record index.

Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        const int n = nums.size();
        if(n < 1) return 0;
        int count = 0, i = 0;
        while(i < n){
            nums[count++] = nums[i];
            int j = i + 1;
            while(j< n && nums[j] == nums[i]) j++;
            i = j;
        }
        return count;
    }
};