怎样找第二大的数,python
发布网友
发布时间：2022-05-05 18:16
共3个回答
热心网友
时间：2022-04-18 10:00
比如说：num_list = [98,12,45,1,2,32,90,45,23,121,11]
直接排序，输出倒数第二个数即可。
num_list = [98,12,45,1,2,32,90,45,23,121,11]
tmp_list = sorted(num_list)
print('第二大的数是：',tmp_list[-2])
扩展资料：
# 最基础的遍历无序列表的查找算法
# 时间复杂度O(n)
def sequential_search(lis, key):
length = len(lis)
for i in range(length):
if lis[i] == key:
return i
else:
return False
if __name__ == '__main__':
LIST = [1, 5, 8, 123, 22, 54, 7, 99, 300, 222]
result = sequential_search(LIST, 123)
print(result)
热心网友
时间：2022-04-18 11:18
sorted(num_list)[-2]
# 方法二，用标准库中堆模块中的nlargest方法
>>> import heapq
>>> num_list = [10,1,2,3,4,5,6,7,8,9,0]
>>> heapq.nlargest(2, num_list)
[10, 9]
>>>
# 方法三，如果不准用内置方法，自己写个算法。
def second_large(num_list):
"""find second largest num in a list.
>>> second_large([1,2,0])
1
>>> second_large([1,2,3,0])
2
>>> second_large([1,3,2,0])
2
>>> second_large([10,1,2,3,4,5,6,7,8,9,0])
9
>>> second_large([1,9,2,3,4,5,6,7,8,10,0])
9"""
iterator = iter(num_list)
largest, second = 0, 0
while True:
num = next(iterator)
if num == 0:
break
elif num > largest:
largest, second = num, largest
elif num > second:
largest, second = largest, num
return second
if __name__ == '__main__':
import doctest
doctest.testmod()
热心网友
时间：2022-04-18 12:53
