Lintcode: remove duplicates from sorted array

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 in place with constant memory.

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Yes

Example

Given input array A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

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Two Pointers Array

Use vector.erase() function:
Just check repetition one by one. Well, just write it for fun.
Time: O(n2)
Space: O(n)

class Solution {
public:
    /**
     * @param A: a list of integers
     * @return : return an integer
     */
    int removeDuplicates(vector<int> &nums) {
        // write your code here
        //check element after current element
        for(int i = 0; i < nums.size(); i++){
            if(i >= nums.size() - 1) break;
            if(nums[i] == nums[i+1]) {
                nums.erase(nums.begin() + i);
                i--;
            }
        }
        return nums.size();
        //check element before current element
        for(int i = 1; i < nums.size(); i++){
            if(i >= nums.size()) break;
            if(nums[i] == nums[i - 1]) {
                nums.erase(nums.begin() + i);
                i--;
            }
        }
        return nums.size();
    }
};

Two pointers:
One pointer iterates each element from the array. The other pointer is used to store element which is not duplicated with others.
Time: O(n)
Space: O(n)

class Solution {
public:
    /**
     * @param A: a list of integers
     * @return : return an integer
     */
    int removeDuplicates(vector<int> &nums) {

        //compare to rear element
       if(nums.size() == 0) return 0;
       int j = 1;
       for(int i = 1; i < nums.size(); i++){
           if(nums[i] != nums[i - 1]) nums[j++] = nums[i];
       }
       return j;
       
       //compare to front element
       if(nums.size() == 0) return 0;
       int j = 0;
       for(int i = 0; i < nums.size() - 1; i++){
           if(nums[i] != nums[i + 1]) nums[j++] = nums[i];
       }
       nums[j] = nums.back();
       return j + 1;
    }
};

Second SHUA:

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class Solution { public: /** * @param A: a list of integers * @return : return an integer */ int removeDuplicates(vector<int> &nums) { // write your code here if(nums.size() == 0) return 0; int cur = 1; for(int i = 1; i < nums.size(); i++){ if(nums[i] != nums[i - 1]){ nums[cur] = nums[i]; cur++; } } return cur; } };