题目： 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].
这道题和上一道题目比较像：Leetcode: Remove Element 都是通过定义一个伪指针，，这个指针记录满足要求的数据位置，当前数据满足要求的时候（不用删除的时候）指针移动一位，最后返回这个伪指针的值。
C++参考代码：
class Solution{public:int removeDuplicates(int A[], int n){if (n == 0) return 0;int pt = 1;for (int i = 1; i < n; i++){if (A[i – 1] != A[i]){A[pt++] = A[i];}}return pt;}};
C#参考代码：
public class Solution{(int[] A){if (A.Length == 0) return 0;int pt = 1;for (int i = 1; i < A.Length; i++){if (A[i – 1] != A[i]) A[pt++] = A[i];}return pt;}}
Python参考代码：
:count = len(A)if count == 0:return 0pt = 1for i in range(1, count):if A[i – 1] != A[i]:A[pt] = A[i]pt += 1return pt
Java参考代码：
public class Solution {(int[] A) {if (A.length == 0) return 0;int pt = 1;for (int i = 1; i < A.length; i++) {if (A[i – 1] != A[i]) {A[pt++] = A[i];}}return pt;}}
最有效的资本是我们的信誉，它24小时不停为我们工作。
