Python describe LeetCode 80. Remove duplicate items from an ordered array II

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Here's an ordered array nums , Would you please ** In situ ** Delete duplicate elements , Elements that appear more than twice Only twice , 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 .

explain ：

Why is the return value an integer , But the output answer is array ？

Please note that , The input array is based on **「 quote 」** By way of transmission , This means that modifying the input array in a function is visible to the caller .

You can imagine the internal operation as follows :

// nums In order to “ quote ” By way of transmission . in other words , Do not make any copies of the arguments
int len = removeDuplicates(nums);
// Modifying the input array in a function is visible to the caller .
// Based on the length returned by your function , It prints out the array Within this length All elements of .
for (int i = 0; i < len; i++) {
print(nums[i]);
}

Example 1：

 Input ：nums = [1,1,1,2,2,3]
Output ：5, nums = [1,1,2,2,3]
explain ： Function should return the new length length = 5, And the first five elements of the original array are modified to 1, 1, 2, 2, 3 . You don't need to think about the elements in the array that follow the new length .

Example 2：

 Input ：nums = [0,0,1,1,1,1,2,3,3]
Output ：7, nums = [0,0,1,1,2,3,3]
explain ： Function should return the new length length = 7, And the first five elements of the original array are modified to 0, 0, 1, 1, 2, 3, 3 . You don't need to think about the elements in the array that follow the new length .

Tips ：

• 1 <= nums.length <= 3 * 104
• -104 <= nums[i] <= 104
• nums In ascending order

Python describe

Double pointer , If there are duplicates , Keep the first two , Then walk straight back , Until different numbers

class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
n = len(nums)
if n == 1:
return n
l,r = 0,0
while r < n:
nums[l] = nums[r]
l,r = l+1,r+1
if r < n and nums[l-1] == nums[r]:
nums[l] = nums[r]
r += 1
while r < n and nums[r] == nums[l]:
r += 1
l += 1
return l