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.

       题目大意：计算一个排好序的数组中不重复元素的个数n，并将个数返回给调用函数，但是同时需要改变数组中元素的排列顺序（不能重新申请内存存储这些数据），最终比较的结果是数组的前n个元素，前n个元素就是n个排好序的数。

样例

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:

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.

python解法

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        num = sorted(list(set(nums)))
        for i, n in enumerate(num):
            nums[i] = n
        return len(num)
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Runtime: 96 ms, faster than 76.82% of Python3 online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 15.6 MB, less than 5.01% of Python3 online submissions for Remove Duplicates from Sorted Array.

题后反思：

1. 列表使用集合去重时数据会变得无序，所以转换成列表需要进行排序

C语言解法

int removeDuplicates(int* nums, int numsSize){
    int n = 1 ;
    if(numsSize == 0)
    {
        return 0;
    }
    for(int i = 1; i < numsSize;i++)
    {
        if(nums[i-1] != nums[i])
        {
            nums[n ++] = nums[i];
        }
    }
    return n;
}
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Runtime: 20 ms, faster than 69.83% of C online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 9.6 MB, less than 31.67% of C online submissions for Remove Duplicates from Sorted Array.

题后反思：

1. 空数组的情况要考虑到

文中都是我个人的理解，如有错误的地方欢迎下方评论告诉我，我及时更正，大家共同进步