2024.5.25AcWing刷题记录-排序篇

一、786. 第k个数 - AcWing题库

三路快速排序

import random
def func(nums, start, end):
    if start >= end:
        return 
    idx = random.randint(start, end)
    base = nums[idx]
    i, j, m = start, start, end + 1
    while j < m:
        if nums[j] < base:
            nums[i], nums[j] = nums[j], nums[i]
            i += 1
            j += 1
        elif nums[j] > base:
            nums[j], nums[m - 1] = nums[m - 1], nums[j]
            m -= 1
        else:
            j += 1
    func(nums, start, i - 1)
    func(nums, m, end)
    
n, k = map(int, input().split())
nums = list(map(int, input().split()))
func(nums, 0, len(nums) - 1)
print(nums[k - 1])

二、787. 归并排序 - AcWing题库

递归归并排序

def merge_sort(nums):
    n = len(nums)
    if n <= 1:
        return nums
    mid = n // 2
    left = merge_sort(nums[:mid])
    right = merge_sort(nums[mid:])
    
    res = []
    i, j = 0, 0
    # 左右列表长度
    while i < mid and j < n - mid:
        if right[j] < left[i]:
            res.append(right[j])
            j += 1
        else:
            res.append(left[i])
            i += 1
    # 将剩下的添加到末尾
    res += left[i:]
    res += right[j:]
    return res
n = int(input())
nums = list(map(int, input().split()))
nums = merge_sort(nums)
for x in nums:
    print(x, end = ' ')

三、788. 逆序对的数量 - AcWing题库

归并排序时计数

# 使用归并排序时计数
def merge_inversions(nums):
    n = len(nums)
    if n <= 1:
        return 0, nums
    mid = n // 2
    cnt_l, left = merge_inversions(nums[:mid])
    cnt_r, right = merge_inversions(nums[mid:])
    i, j = 0, 0
    res = []
    cnt = 0
    while i < mid and j < n - mid:
        if left[i] > right[j]:
            res.append(right[j])
            j += 1
            cnt += mid - i  # 注意这里，是left从i到结尾都是算逆序
        else:
            res.append(left[i])
            i += 1
    res += left[i:]
    res += right[j:]
    # print(cnt, cnt_l, cnt_r)
    return cnt + cnt_l + cnt_r, res
n = int(input())
nums = list(map(int, input().split()))
cnt, _ = merge_inversions(nums)
print(cnt)

完

感谢你看到这里！一起加油吧！