LeetCode26——Remove Duplicates from Sorted Array

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_28306361/article/details/87797175

题目链接

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.

代码

class Solution:
    def removeDuplicates(self, nums: 'List[int]') -> 'int':
        if len(nums) == 1:
            return 1
        if len(nums) == 0:
            return 0
        length = 1
        i = 1
        while i < len(nums):
            if nums[i] == nums[length-1]:
                i += 1
            elif nums[i] != nums[length-1]:
                length += 1
                nums[length-1] = nums[i]
                i += 1

        return length

简单题型，，由于给定的是一个有序数组，设置两个指针，一个指向不重复的最后一个位置，一个往后扫描就可以了。