【剑指Offer】 29.最小的K个数 python实现

题目描述

输入n个整数,找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字,则最小的4个数字是1,2,3,4,。


方法一:蒂姆排序


# -*- coding:utf-8 -*-
class Solution:
    def GetLeastNumbers_Solution(self, tinput, k):
        # write code here
        a = sorted(tinput)
        if len(a)<k:
            return []
        else:
            return a[:k]

方法二:快速排序


# -*- coding:utf-8 -*-
class Solution:
    def GetLeastNumbers_Solution(self, tinput, k):
        # write code here
        def quick_sort(lst):
            if not lst:
                return []
            pivot = lst[0]
            left = quick_sort([x for x in lst[1: ] if x < pivot])
            right = quick_sort([x for x in lst[1: ] if x >= pivot])
            return left + [pivot] + right
 
        if tinput == [] or k > len(tinput):
            return []
        tinput = quick_sort(tinput)
        return tinput[: k]

方法三:归并排序


# -*- coding:utf-8 -*-
class Solution:
    def GetLeastNumbers_Solution(self, tinput, k):
        # write code here
        def merge_sort(lst):
            if len(lst) <= 1:
                return lst
            mid = len(lst) // 2
            left = merge_sort(lst[: mid])
            right = merge_sort(lst[mid:])
            return merge(left, right)
        def merge(left, right):
            l, r, res = 0, 0, []
            while l < len(left) and r < len(right):
                if left[l] <= right[r]:
                    res.append(left[l])
                    l += 1
                else:
                    res.append(right[r])
                    r += 1
            res += left[l:]
            res += right[r:]
            return res
        if tinput == [] or k > len(tinput):
            return []
        tinput = merge_sort(tinput)
        return tinput[: k]

方法四:堆排序


# -*- coding:utf-8 -*-
class Solution:
    def GetLeastNumbers_Solution(self, tinput, k):
        # write code here
        def siftup(lst, temp, begin, end):
            if lst == []:
                return []
            i, j = begin, begin * 2 + 1
            while j < end:
                if j + 1 < end and lst[j + 1] > lst[j]:
                    j += 1
                elif temp > lst[j]:
                    break
                else:
                    lst[i] = lst[j]
                    i, j = j, 2 * j + 1
            lst[i] = temp
 
        def heap_sort(lst):
            if lst == []:
                return []
            end = len(lst)
            for i in range((end // 2) - 1, -1, -1):
                siftup(lst, lst[i], i, end)
            for i in range(end - 1, 0, -1):
                temp = lst[i]
                lst[i] = lst[0]
                siftup(lst, temp, 0, i)
            return lst
 
        if tinput == [] or k > len(tinput):
            return []
        tinput = heap_sort(tinput)
        return tinput[: k]

方法五:冒泡排序


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方法六:直接选择排序


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方法七:插入排序


在这里插入代码片
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