Python把数组排成最小数

2020-01-21T15:46:18

首先将数组中的数字全部转换为字符串存储在一个新的数组中,然后比较每两个数字串的拼接的mn和nm的大小,若mn< nm,则m更小,反之n更小,然后把更小的数放入一个新的List中,最后输出即可。使用冒泡排序很方便。

'''
输入一个正整数数组,把数组里所有数字拼接起来排成一个数,打印能拼接出的所有数字中最小的一个。
例如输入数组{3,32,321},则打印出这三个数字能排成的最小数字为321323。
'''
# 运行环境Python 3.x
from functools import cmp_to_key
# -*- coding:utf-8 -*-
class Solution:
    def PrintMinNumber(self, numbers):
        if numbers == None or len(numbers) <= 0:
            return ''
        strList = []
        for i in numbers:
            strList.append(str(i))
        # key是一种比较规则
        # 比较 x+y 和 x-y 的大小, 因为为str型, 需要先转换成int型
        key = cmp_to_key(lambda x, y: int(x+y)-int(y+x))
        strList.sort(key=key)
        return ''.join(strList)
    # 使用冒泡排序
    def PrintMinNumber2(self, numbers):
        if numbers == None or len(numbers) <= 0:
            return ''
        strNum = [str(m) for m in numbers]
        for i in range(len(numbers)-1):
            for j in range(i+1, len(numbers)):
                if strNum[i] + strNum[j] > strNum[j] + strNum[i]:
                    strNum[i], strNum[j] = strNum[j], strNum[i]
        return ''.join(strNum)

numbers = [3, 32, 321]
s = Solution()
print(s.PrintMinNumber(numbers))

# 运行环境 Python 2.7

# class Solution:
#     def PrintMinNumber(self, numbers):
#         # write code here
#         if numbers == None or len(numbers) == 0:
#             return ""
#         A = ["" for i in range(len(numbers))]
#         for i in range(len(numbers)):
#             A[i] = str(numbers[i])
#         A.sort(self.cmp)
#         return "".join(A)
#     def cmp(self,e1,e2):
#         s1 =e1+e2
#         s2 = e2 + e1
#         return cmp(s1,s2)
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