feat: 26.remove-duplicates-from-sorted-array add Python3 implementation

problems/26.remove-duplicates-from-sorted-array.md

@@ -67,6 +67,9 @@ for (int i = 0; i < len; i++) {
 
 ## 代码
 
+* 语言支持：JS，Python
+
+Javascript Code:
 ```js
 /*
  * @lc app=leetcode id=26 lang=javascript
@@ -83,51 +86,51 @@ for (int i = 0; i < len; i++) {
  *
  * Given a sorted array nums, 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 by modifying
  * the input array in-place with O(1) extra memory.
- * 
+ *
  * Example 1:
- * 
- * 
+ *
+ *
  * Given 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 returned length.
- * 
+ *
  * Example 2:
- * 
- * 
+ *
+ *
  * Given nums = [0,0,1,1,1,2,2,3,3,4],
- * 
+ *
  * Your function should return length = 5, with the first five elements of nums
  * being modified to 0, 1, 2, 3, and 4 respectively.
- * 
+ *
  * It doesn't matter what values are set beyond the returned length.
- * 
- * 
+ *
+ *
  * Clarification:
- * 
+ *
  * Confused why the returned value is an integer but your answer is an array?
- * 
+ *
  * Note that the input array is passed in by reference, which means
  * modification to the input array will be known to the caller as well.
- * 
+ *
  * Internally you can think of this:
- * 
- * 
+ *
+ *
  * // 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]);
  * }
- * 
+ *
  */
 /**
  * @param {number[]} nums
@@ -140,8 +143,23 @@ var removeDuplicates = function(nums) {
  if (nums[fastP] !== nums[slowP]) {
  slowP++;
  nums[slowP] = nums[fastP]
- } 
+ }
  }
  return slowP + 1;
 };
 ```
+
+Python Code:
+```python
+class Solution:
+ def removeDuplicates(self, nums: List[int]) -> int:
+ if nums:
+ slow = 0
+ for fast in range(1, len(nums)):
+ if nums[fast] != nums[slow]:
+ slow += 1
+ nums[slow] = nums[fast]
+ return slow + 1
+ else:
+ return 0
+```