[LeetCode] 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].

最笨的方法，从前往后，发现一个重复的数字，长度减1，把数组中重复数字后面的数依次前移覆盖重复的数字。

LeetCode结果：Time Limit Exceeded,编码如下：

int removeDuplicates(int A[], int n) {    
        if(n<=1)
           return n; 
        int len = n;
        for(int i=1;i<len;){
            if(A[i]==A[i-1]){
                if(i!=len-1)
                   for(int j = i;j<len-1;j++)
                      A[j]=A[j+1];
                len--; 
            }else{
            i++;}
        }//end for
        return len;
    }

 用时间复杂度是O(n)的方法做，即：在原来的数组中，遍历一次，逐渐将之前重复的取掉，不重复的挪到前面本该存放的地方。

  Accept编码如下：

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        int len = n;
        int newIndex = 0,oldIndex;
        for(oldIndex = 1;oldIndex<n;oldIndex++){
            if(A[oldIndex]==A[newIndex])
                len--;
            else{
                newIndex++;
                if(newIndex != oldIndex)
                   A[newIndex]=A[oldIndex];
            }//end if
        }//end for
        return len;
    }
};