[leetcode:python]26.Remove Duplicates from Sorted Array

最新推荐文章于 2019-06-18 12:02:54 发布

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最新推荐文章于 2019-06-18 12:02:54 发布

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分类专栏：算法文章标签： leetcode python

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本文介绍了一种不使用额外空间的方法来移除有序数组中的重复元素，并保持元素的相对顺序不变。通过两个不同的实现方式，展示了如何有效地将数组中重复的元素压缩到前部，同时返回新的长度。

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题目：
Given a sorted array, 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 in place with constant memory.

For example,
Given input array 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 new length.
题意：
给定一个有序数组，移除重复的元素，返回新的长度
不要分配额外空间给新的数组，空间复杂度为O（1）

方法一：性能82ms

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
            return 0
        j = 0
        for i in range(len(nums)):
            if nums[i] != nums[j]:
                nums[i], nums[j+1] = nums[j+1], nums[i]
                j = j+1
        return j+1

方法二：性能75ms

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
            return None
        current = nums[0]
        same = 0
        for index in range(1, len(nums)):
            if nums[index] == current:
                same += 1
            else:
                nums[index-same] = nums[index]
                current = nums[index]

        return len(nums)-same