@@ -67,6 +67,9 @@ for (int i = 0; i < len; i++) {
6767
6868## 代码
6969
70+ * 语言支持:JS,Python
71+ 72+ Javascript Code:
7073``` js
7174/*
7275 * @lc app=leetcode id=26 lang=javascript
@@ -83,51 +86,51 @@ for (int i = 0; i < len; i++) {
8386 *
8487 * Given a sorted array nums, remove the duplicates in-place such that each
8588 * element appear only once and return the new length.
86- *
89+ *
8790 * Do not allocate extra space for another array, you must do this by modifying
8891 * the input array in-place with O(1) extra memory.
89- *
92+ *
9093 * Example 1:
91- *
92- *
94+ *
95+ *
9396 * Given nums = [1,1,2],
94- *
97+ *
9598 * Your function should return length = 2, with the first two elements of nums
9699 * being 1 and 2 respectively.
97- *
100+ *
98101 * It doesn't matter what you leave beyond the returned length.
99- *
102+ *
100103 * Example 2:
101- *
102- *
104+ *
105+ *
103106 * Given nums = [0,0,1,1,1,2,2,3,3,4],
104- *
107+ *
105108 * Your function should return length = 5, with the first five elements of nums
106109 * being modified to 0, 1, 2, 3, and 4 respectively.
107- *
110+ *
108111 * It doesn't matter what values are set beyond the returned length.
109- *
110- *
112+ *
113+ *
111114 * Clarification:
112- *
115+ *
113116 * Confused why the returned value is an integer but your answer is an array?
114- *
117+ *
115118 * Note that the input array is passed in by reference, which means
116119 * modification to the input array will be known to the caller as well.
117- *
120+ *
118121 * Internally you can think of this:
119- *
120- *
122+ *
123+ *
121124 * // nums is passed in by reference. (i.e., without making a copy)
122125 * int len = removeDuplicates(nums);
123- *
126+ *
124127 * // any modification to nums in your function would be known by the caller.
125128 * // using the length returned by your function, it prints the first len
126129 * elements.
127130 * for (int i = 0; i < len; i++) {
128131 * print(nums[i]);
129132 * }
130- *
133+ *
131134 */
132135/**
133136 * @param {number[]} nums
@@ -140,8 +143,23 @@ var removeDuplicates = function(nums) {
140143 if (nums[fastP] !== nums[slowP]) {
141144 slowP++ ;
142145 nums[slowP] = nums[fastP]
143- }
146+ }
144147 }
145148 return slowP + 1 ;
146149};
147150```
151+ 152+ Python Code:
153+ ``` python
154+ class Solution :
155+ def removeDuplicates (self nums : List[int ]) -> int :
156+ if nums:
157+ slow = 0
158+ for fast in range (1 , len (nums)):
159+ if nums[fast] != nums[slow]:
160+ slow += 1
161+ nums[slow] = nums[fast]
162+ return slow + 1
163+ else :
164+ return 0
165+ ```
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