Leetcode: Remove Element

称号：
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.

这道题比較简单直接看代码，这里多写几个版本号的作參考！

C++版本号

class Solution
{
public:
    int removeElement(int A[], int n, int elem)
    {
        int pt = 0;
        for (int i = 0; i < n; i++)
        {
            if (A[i] != elem) A[pt++] = A[i];
        }
        return pt;
    }
};

C#版本号：

public class Solution
{
    public int RemoveElement(int[] A, int elem)
    {
        int pt = 0;
        for (int i = 0; i < A.Length; i++)
        {
            if (A[i] != elem) A[pt++] = A[i];
        }
        return pt;
    }
}

Python版本号：

class Solution:
    # @param    A       a list of integers
    # @param    elem    an integer, value need to be removed
    # @return an integer
    def removeElement(self, A, elem):
        pt = 0
        for i in range(len(A)):
            if A[i] != elem:
              A[pt] = A[i]
              pt += 1
        return pt

Java版本号：

public class Solution
{
    public int removeElement(int[] A, int elem) {
        int pt = 0;
        for (int i = 0; i < A.length; i++)
        {
            if (A[i] != elem) A[pt++] = A[i];
        }
        return pt;
    }
}

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