// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
  print(nums[i]);
}

def removeDuplicates(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    if not nums:
      return 0
    if len(nums)==1:  #單獨(dú)判斷列表長(zhǎng)度為1的情況，因?yàn)橹蟮膄or循環(huán)從下標(biāo)1開始
      return 1
    temp_num = nums[0]
    count =0     #for循環(huán)中動(dòng)態(tài)刪除列表元素，列表縮短，為了防止下標(biāo)溢出需要用count標(biāo)記刪除元素個(gè)數(shù)
    for index, num in enumerate(nums[1:]):
      if temp_num == num:   #元素相等就刪除
        del nums[index-count]
        count += 1
      else:
        temp_num = num
    return len(nums)
def removeDuplicates(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    forth = 0
    back = 1
    while back <= len(nums)-1:
      if nums[forth] == nums[back]:
        nums.pop(back)
      else:
        forth += 1
        back += 1
    return len(nums)

def removeDuplicates(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    if not nums:
      return 0
    length = 0   #不存在重復(fù)數(shù)字的數(shù)組長(zhǎng)度
    for index in range(1,len(nums)):   #遍歷數(shù)組
      if nums[index] != nums[length]:
        length += 1
        nums[length] = nums[index]
    return length+1