Leetcode 每日一题
题目链接: 922. 按奇偶排序数组 II
难度: 简单
解题思路: 先桶排序一下,可以统计每个元素的出现的次数和他的奇偶性。然后再按照奇偶性填入原数组。
题解:

class Solution:
    def sortArrayByParityII(self, A: List[int]) -> List[int]:
        length = len(A)
        bucket = [0] * 1001
        bucket_len = 0

        # 桶排序
        for example in A:
            bucket[example] += 1
            bucket_len = max(bucket_len, example)

        bucket_index = 0
        A_index = 0
        # 偶数
        while A_index < length and bucket_index <= bucket_len:
            if bucket[bucket_index] != 0:
                A[A_index] = bucket_index
                bucket[bucket_index] -= 1
                A_index += 2
            else:
                bucket_index += 2
        
        bucket_index = 1
        A_index = 1
        # 奇数
        while A_index < length and bucket_index <= bucket_len:
            if bucket[bucket_index] != 0:
                A[A_index] = bucket_index
                bucket[bucket_index] -= 1
                A_index += 2
            else:
                bucket_index += 2
        
        return A