从零开始刷力扣——统计数组中的元素

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        count = {}
        duplicate = -1
        missing = -1
        
        # 记录每个数字的出现次数
        for num in nums:
            if num not in count:
                count[num] = 1
            else:
                count[num] += 1
                
        # 找到重复的数字和丢失的数字
        for i in range(1, len(nums) + 1):
            if i not in count:
                missing = i
            elif count[i] > 1:
                duplicate = i
        
        return [duplicate, missing]

class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        # 计算数组的度
        counter = {}
        for num in nums:
            if num in counter:
                counter[num] += 1
            else:
                counter[num] = 1
        max_degree = max(counter.values())

        # 找到数组中出现频数为度的数字的第一次出现位置和最后一次出现位置
        first_occurrence = {}
        last_occurrence = {}
        for i, num in enumerate(nums):
            if num not in first_occurrence:
                first_occurrence[num] = i
            last_occurrence[num] = i

        # 计算每个数字的子数组长度
        min_length = len(nums)
        for num, count in counter.items():
            if count == max_degree:
                min_length = min(min_length, last_occurrence[num] - first_occurrence[num] + 1)

        return min_length

class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        n = len(nums)
        # 使用HashSet记录数组中出现的数字
        num_set = set(nums)
        # 存放缺失的数字
        result = []

        # 遍历 [1, n] 范围内的数字
        for i in range(1, n+1):
            # 如果当前数字不在HashSet中，说明缺失
            if i not in num_set:
                result.append(i)

        return result

class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        n = len(nums)

        # 第一次遍历，将出现的数字对应的索引位置标记为负数
        for num in nums:
            index = abs(num) - 1
            nums[index] = -abs(nums[index])

        # 第二次遍历，找出非负数的索引位置，即为缺失的数字
        result = [i + 1 for i in range(n) if nums[i] > 0]

        return result

class Solution:
    def findDuplicates(self, nums: List[int]) -> List[int]:
        ans = []
        for num in nums:
            if nums[abs(num) - 1] < 0:
                ans.append(abs(num))
            nums[abs(num) - 1] *= - 1
        return ans