Leetcode 26 Remove Duplicates from Sorted Array

Remove Duplicates from Sorted Array"/>

Leetcode 26 Remove Duplicates from Sorted Array

Leetcode 26 Remove Duplicates from Sorted Array

题目描述

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]);
}

来源：力扣（LeetCode）
链接：
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路解析

• 该题的关键是要求空间复杂度为 O ( 1 ) O(1) O(1)，也就是要求我们只在原列表上进行操作，不能新建列表。

class Solution:# 双指针法def removeDuplicates(self, nums: List[int]) -> int:if nums == []:return 0i = 0length = len(nums)for j in range(0, length):if nums[i] == nums[j]:j += 1else:i += 1nums[i] = nums[j]                return i + 1

这里提到了一类算法，in-place algorithm