leetcode

class :
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        d = {}
        ans = []
        for i in range(1, len(nums)+1):
            d.setdefault(i, 0)
        for num in nums:
            d[num] += 1
        for i in range(1, len(nums)+1):
            if (d[i] == 0):
                ans.append(i)
        return ans

# O(N)时间复杂度，且最终返回的list不计入额外存储空间，因此是O(0)的空间复杂度
class :
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        r = [*range(1, len(nums) + 1)]
        for n in nums: r[n - 1] = 0
        for i in range(len(r) - 1, -1, -1):
            if not r[i]: r.pop(i)
        return r

# 同上一解法
class :
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        for n in nums:
            nums[abs(n) - 1] = -abs(nums[abs(n) - 1])
        return [i + 1 for i, n in enumerate(nums) if n > 0]

# 使用集合，很快，但是空间复杂度应该超过要求了
class :
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        uniq = set(nums)
        all_num = set([i for i in range(1, len(nums)+1)])
        return  list(all_num - uniq)