【Leetcode Algorithm】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 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.

自己的代码：
public class Solution {
public int removeDuplicates(int[] nums) {
int len = nums.length;
int i=0;
//逐个遍历长度为len的int数组
while(i<len-1){
//由于数组是排序的，所以接下来需要计算出某元素k与其后面的元素相同的数量count
int k = i;
int count = 0;
while(k+1<len){
if(nums[k++]==nums[k]){
count++;
}
else{
break;
}
}
//当有重复时，把剩余的元素覆盖重复的元素
if(count>0){
int j = i;
while(++j+count<len){
nums[j] = nums[j+count];
}
//更新数组的长度
len -= count;
}
i++;
}
return len;
}
}
虽然accepted了，但是用时696ms，不够好
当有重复时，没有必要把剩余的所有元素都往左移

改进的代码：

public class Solution {
public int removeDuplicates(int[] nums) {
//边界条件
if (nums.length < 2)
return nums.length;

int j = 0;
int i = 1;
//逐个遍历nums数组
while (i < nums.length) {
//使用两个指针i,j，当索引为i,j的元素不同时，覆盖一次
if (nums[i] != nums[j]) {
nums[++j] = nums[i];
}
i++;
}
return j + 1;
}
}