【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].

【题意】

对一个有序的数组去重，不适用额外的空间，返回去重后的数组长度

【思路】

【代码】

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n==0||n==1)return n;
        int p2=0;   //指向新数组尾节点的游标
        int p1=1;   //元数组游标
        while(p1<n){
            if(A[p1]==A[p2])p1++;
            else{
                A[p2+1]=A[p1];
                p2++;p1++;
            }
        }
        return p2+1;
    }
};