36、数组中出现次数超过一半的数字
只会最笨的用字典遍历计数，然后判断value是否大于长度的一半。看了下别人的方法，参考思路来自博客：https://www.cnblogs.com/python27/archive/2011/12/15/2289534.html
自己用python实现。

# -*- coding:utf-8 -*-
class Solution:
    def MoreThanHalfNum_Solution(self, numbers):
        # write code here
        if not numbers:
            return 0
        else:
            cur = [numbers[0], 1]
            for i in numbers[1:]:
                if i == cur[0]:
                    cur[1] += 1
                else:
                    if cur[1] > 1:
                        cur[1] -= 1
                    elif cur[1] == 1:
                        cur = [i, 1]
            print(cur)
            return cur[0] if numbers.count(cur[0])>len(numbers)/2 else 0

37、整数中1出现的次数
多试几个数字就能找到规律了。虽然我也试了很久。写的比较长，但逻辑应该没有什么错误，可能可以在代码上进行改进。

# -*- coding:utf-8 -*-
class Solution:
    def NumberOf1Between1AndN_Solution(self, n):
        temp = str(n)
        nums = [int(x) for x in list(temp)[::-1]]
        count = 0
        for i, num in enumerate(nums[:-1]):
            pos = 10**i
            pre = int(temp[:-(i+1)])
            if num >= 1:
                count += pos
                count += pos*pre
            if num < 1:
                count += pos*pre
        if nums[-1] > 1:
            count += 10**(len(temp)-1)
        else:
            count += int(temp[1:])+1
        return count

38、把数组排成最小的数
不会做，看了别人的思路。自己写了一遍冒泡排序的过程。等看了排序相关内容再重新用快排写一遍

class Solution:
    def PrintMinNumber(self, numbers):
        def swap(i, j):
            numbers[i], numbers[j] = numbers[j], numbers[i]

        def sortbynumadd(numbers):
            flag = True
            i = len(numbers)
            while i > 0 and flag:
                flag = False
                for j in range(i-1):
                    if int(str(numbers[j]) + str(numbers[j+1])) > int(str(numbers[j+1]) + str(numbers[j])):
                        swap(j, j+1)
                        flag = True
                i -= 1
            return numbers

        rst = sortbynumadd(numbers)
        return ''.join([str(x) for x in rst])

39、数组中重复的数字
简单的直接用字典，时间空间复杂度都是O(n)。参考了别人的思路，时间复杂度O(n)空间复杂度O(1)

class Solution:
    def duplicate(self, numbers, duplication):
        long = len(numbers)
        for i in range(len(numbers)):
            index = numbers[i]%long if numbers[i] >= long else numbers[i]
            if numbers[index] > long:
                duplication[0] = index
                return True
            numbers[index] += long
        return False

40、构造乘积数组
还是不会做。。参考https://www.nowcoder.com/questionTerminal/94a4d381a68b47b7a8bed86f2975db46
的第一条评论

class Solution:
    def multiply(self, A):
        B = [1]
        for i in range(len(A)-1):
            B.append(B[i]*A[i])
        temp = 1
        pos = -2
        for j in range(len(A)-1, 0, -1):
            temp *= A[j]
            B[pos] *= temp
            pos -= 1
        return B

感觉今天的五道题大部分都不会做😔。还需要继续努力