[LeetCode] Remove Element、Remove Duplicates from Sorted Array、Remove Duplicates from Sorted Array II

Remove Element:Given an array and a value, remove all instances of that value in place and return the new length.The order of elements can be changed. It doesn't matter what you leave beyond the new length.

class Solution {
public:
    int removeElement(int A[], int n, int elem) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        assert(A&&n>=0);
	    int del=0;
	    for(int i=0;i<n;i++)
	    	if (A[i]==elem)
			del++;
	    	else
			A[i-del]=A[i];
	    return n-del;
    }
};

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.

For example,
Given input array A = [1,1,2],

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

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        assert(A&&n>=0);
	    if (n==0 ) return 0;
	    int del=0;
	    int pre=A[0];
	    for(int i=1;i<n;i++)
	    {
		    if ( A[i]==pre )
			    del++;
		    else
		    {
			    A[i-del]=A[i];
			    pre=A[i];
		    }
	    }
	    return n-del;
    }
};

Remove Duplicates from Sorted Array II

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

class Solution {
public:
    int removeDuplicates(int A[], int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function

		//assert(A&&n>=0);
		if ( n<=2 )
			return n;

		int curCh=A[0];
		int time=1;
		int del=0;
		for(int i=1;i<n;i++)
		{
			if(A[i]==curCh)
			{
				if ( time <2 )
				{
					A[i-del]=A[i];
					time++;
				}
				else
					del++;
			}
			else
			{
				A[i-del]=A[i];
				curCh=A[i];
				time=1;
			}
		}
		return n-del;
	}
};