Force buckle deletes duplicates in the sort array

Here's an ordered array nums , Would you please In situ Delete duplicate elements , Make each element Only once , Returns the new length of the deleted array .

Don't use extra array space , You must be there. In situ Modify input array And using O(1) Complete with extra space .

 Example  1：
 Input ：nums = [1,1,2]
 Output ：2, nums = [1,2]
 explain ： Function should return the new length  2 , And the original array  nums  The first two elements of are modified to  1, 2 . You don't need to think about the elements in the array that follow the new length .
 Example  2：
 Input ：nums = [0,0,1,1,1,2,2,3,3,4]
 Output ：5, nums = [0,1,2,3,4]
 explain ： Function should return the new length  5 ,  And the original array  nums  The first five elements of are modified to  0, 1, 2, 3, 4 . You don't need to think about the elements in the array that follow the new length .
 Tips ：
0 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
nums  In ascending order 

class Solution {<!-- -->
    public int removeDuplicates(int[] nums) {<!-- -->
        //  Special case handling 
        if(nums==null||nums.length==0){<!-- -->
            return 0;
        }
        int left = 0,right = 0;
        while (right&lt;nums.length){<!-- -->
            if(nums[left]!=nums[right]) {<!-- -->
                nums[++left] = nums[right];
            }
            right++;
        }
        return left+1;
    }
}