26. Remove Duplicates from Sorted Array

题目链接

 思路一：单指针法，遍历数组，遇到和下一个元素不同的把这个元素按顺序保存到数组的前面，用指针记录保存的个数，注意数组只有一个元素和遍历到最后俩元素的特殊情况。

class Solution:
    def removeDuplicates(self,nums):
        sum = 0
        if len(nums) <= 1:
            return 1
        for i in range(len(nums)-1):
            if i == (len(nums)-2):
                if nums[i] != nums[i+1]:
                    nums[sum] = nums[i]
                    nums[sum+1] = nums[i+1]
                    sum +=2

                else:
                    nums[sum] = nums[i]
                    sum +=1
            else:
                if nums[i] != nums[i+1]:
                    nums[sum] = nums[i]
                    sum +=1
        nums = nums[:sum]
        return sum
if __name__ == '__main__':#提交时main函数内容不用带上
    nums = [1]
    solution = Solution()
    print(solution.removeDuplicates(nums))

思路二，双指针和思路一大同小异

class Solution:
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) <= 1:
            return len(nums)
        i = 0
        j = 1
        while (j < len(nums)):
            if nums[i] == nums[j]:
                nums.pop(j)
            else:
                j = j + 1
                i = j - 1
        return len(nums)

 思路三：使用set，但是貌似开辟空间了，但是在力扣也通过了

class Solution:
    def removeDuplicates(self, nums):
        nus = list(set(nums))#得到一个无序的无重复数字的list
        nus = sorted(nus)
        for i in range(len(nus)):
            nums[i] = nus[i]
        return len(nums[:len(nus)])