给定一个list 由一些非负整数组成 ，重新排列他们的顺序把他们组成一个最大的整数。
例子：
输入：
[30,1]
返回值：
“301”
def fn(s):
pass
assert fn([30,1]) == "301"
assert fn([1, 201, 20, 9, 8]) == "98202011"
assert fn([1, 203, 20, 9, 8]) == "98203201"
备注：需要保证列表里面的每个元素不被拆开
因重新组合的整数可能较大，将整数变为字符串后返回
题目来源：最大数_牛客题霸_牛客网
import itertools
def fn(list):
ls = []
for p in itertools.permutations(list):
a = ''
for x in p:
a = a + str(x)
ls.append(int(a))
print(str(max(ls)))
return str(max(ls))
assert fn([30,1]) == "301"
assert fn([1, 201, 20, 9, 8]) == "98202011"
assert fn([1, 203, 20, 9, 8]) == "98203201"
这个狠，还有这么个方法
排列组合有点耗性能呀…会出现一些不必要的组合数据，数据量一大会存在问题.
def fn(nums):
# 根据首位先排序
s = list(map(str, nums))
ordered = sorted(s, key=lambda x:str(x), reverse=True)
# 首位相同的，进行二次排序
for i in range(len(ordered)-1):
if ordered[i][0] == ordered[i+1][0]:
if int(ordered[i][-1]) < int(ordered[i+1][0]):
ordered[i], ordered[i+1] = ordered[i+1], ordered[i]
return ''.join(ordered)
assert fn([30,1]) == "301"
assert fn([1, 201, 20, 9, 8]) == "98202011"
assert fn([1, 203, 20, 9, 8]) == "98203201"
import itertools
def fn(nums):
pertmute = list(itertools.permutations([str(x) for x in nums], len(nums)))
result = list(map(lambda x:''.join(x), pertmute))
return max(result)
assert fn([30,1]) == "301"
assert fn([1, 201, 20, 9, 8]) == "98202011"
assert fn([1, 203, 20, 9, 8]) == "98203201"
import itertools
def fn(nums):
total = []
for i in range(10)[::-1]:
data = list(filter(lambda x:str(x).startswith(str(i)), nums))
if data:
if len(data) > 1:
p = list(itertools.permutations(data, len(data)))
total.append([max([str(x)+str(y) for x,y in p])])
else:
total.append(data)
return ''.join([str(x[0]) for x in total])
assert fn([30,1]) == "301"
assert fn([1, 201, 20, 9, 8]) == "98202011"
assert fn([1, 203, 20, 9, 8]) == "98203201"
