Implementación de varias funciones comunes de Python en las preguntas de la entrevista (actualización a largo plazo)

1, función de partición y encuentra el número en la matriz

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        len1 = len(nums1)
        len2 = len(nums2)
        if len1 == 0:
            if len2 & 1 == 0:#偶数
                return (nums2[len2//2]+nums2[(len2//2)-1])/2
            else:
                return nums2[len2//2]
        if len2 == 0:
            if len1 & 1 == 0:#偶数
                return (nums1[len1//2]+nums1[(len1//2)-1])/2
            else:
                return nums1[len1//2]
        sum_len = len1 + len2
        if nums1[-1] <= nums2[0] or nums2[-1] <= nums1[0]:#list1的最后一个都没有list2的第一个大
            sum_list = nums1 + nums2 if (nums1[-1] <= nums2[0]) else (nums2 + nums1)
            if sum_len & 1 == 1:#odd
                return sum_list[sum_len//2]
            else:
                return (sum_list[sum_len//2] + sum_list[(sum_len//2)-1])/2

        sum_list = nums1 + nums2 
        print(sum_list)
        if sum_len & 1 == 1:#奇数长度，取sum_len//2
            val,_ = self.GetKthxiao(sum_list,sum_len//2)
            return val#self.GetKthxiao(sum_list,sum_len//2)
        elif sum_len & 1 == 0:#偶数长度
            v1,paixu1 = self.GetKthxiao(sum_list,(sum_len//2)-1)
            v2,paixu2 = self.GetKthxiao(paixu1[(sum_len//2):],0)
            return (v1+v2)/2#(self.GetKthxiao(sum_list,sum_len//2) + self.GetKthxiao(sum_list,(sum_len//2)-1))/2
        
        return sum_list[sum_len//2]
    def swap(self,inputlist,s1,s2):
        temp = inputlist[s1]
        inputlist[s1] = inputlist[s2]
        inputlist[s2] = temp
        return inputlist

    def Partition(self,inputlist,start,end,value):
        #这个函数把inputlist里比value大的元素放其右边，把比value小的元素放其左边
        #定义：指针cur:表示当前的游标
        #定义：指针small：表示前small个数确保比value小。初值：start - 1
        #定义：指针big：表示value及其后面的数字确保比value大。初值：end + 1
        #cur从start开始

        small = start - 1
        big = end + 1
        cur = start
        length = len(inputlist)

        while cur != big:
            if inputlist[cur] > value:#当前数字大于mid数字
                big -= 1
                inputlist = self.swap(inputlist,cur,big)
            elif inputlist[cur] < value:#当前数字小于mid数字
                small += 1
                inputlist = self.swap(inputlist,cur,small)
                cur += 1
            else:
                cur += 1
        return inputlist,small+1
    def GetKthxiao(self,inputlist,K):
        length = len(inputlist)
        inputlist,index = self.Partition(inputlist,0,length-1,inputlist[0])
        #代表inputlist[:index]都比inputlist[index]小
        while True:
            print(inputlist)
            if index < K:
                inputlist,index = self.Partition(inputlist,index+1,length-1,inputlist[index+1])
            elif index > K:
                inputlist,index = self.Partition(inputlist,0,index-1,inputlist[index-1])
            elif index == K:
                #代表inputlist[:K]都比inputlist[K]小
                #inputlist[K]是inputlist第K小的数字
                return inputlist[K],inputlist

2. Genere una disposición completa de todos los elementos en una matriz

#生成A=[1,1,2,3]的全排列
#递归实现
alllist = []
def quanpailie(inputlist,res=''): 
    for i,a in enumerate(inputlist):
        tempres = res#.copy()
        tempres += str(a)
        templist = []
        for j in range(len(inputlist)):
            if j != i:
                templist.append(inputlist[j])
        if len(templist) > 0:
            tempres += ' '
            quanpailie(templist,tempres)
        else:
            alllist.append(tempres)
quanpailie([1,1,2,3])
print(list(set(alllist)))