Remove Duplicates from Sorted Array

1. Question

给定有序数组，去掉其中的重复元素，使得每个元素仅出现一次。要求实现是in place的，即仅能使用常数级的的额外空间。

要求返回新数组的长度，同时原数组的该长度内是要求的数，该长度以后的数组内容无所谓。

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 nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

2. Solution

考虑以下特殊情况：

数组长度 < 2：不可能存在重复字符

public class Solution {
public int removeDuplicates(int[] A){
int len = A.length;
if( len==0 || len==1 )    return len;

int rLen = 1;   //the new ArrayLength. It has one element at least, the first element.
int p = 1;  //pointer
while( p<len ){
//if present is the same as the previous, p++
if( A[p] == A[p-1] ){
p++;
continue;
}
//present is one of the new array elements
if( p!=rLen )
A[rLen] = A[p];
rLen++;
p++;
}
return rLen;
}
}

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