Python 面試題：輸入一個陣列，輸出該陣列的第二大的數字

問題：
輸入一個陣列，輸出該陣列的第二大的數字，并且撰寫相關的測驗用例

注意：

1.如果list含有非int, float元素需要remove
2.如果list有重復的最大元素，需要自己處理，內置的list.sort(reverse=True)和heapq.nlargest排序，元素個數不變，

附上代碼

removeInvalidItems 去掉不是int或float型別的值，
注意：不能像下邊這樣用一次回圈，因為remove某個元素，下標發生了改變，有些值并不能移除

for item in l:  # remove non_value items
    if not isinstance(item, (int, float)):
        l.remove(item)

下邊是可用代碼，檔案名為 findSecondUtil.py

def removeInvalidItems(l):
    tmpl = list()
    for item in l:
        if not isinstance(item, (int, float)):
            tmpl.append(item)
    for item in tmpl:
        l.remove(item)
    return l

findSecondItem.py 實作找到第二大數字的第一種方法，這種方法不用去掉重復元素，

import sys
from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems
#from findSecondUtil import removeInvalidItems
 
def findSecond(l):
    l = removeInvalidItems(l)
    length = len(l)
    if length == 0:
        print("there is no number item in the list")
        return None
    elif length == 1:
        print("there is only one number item, it's ", l[0])
        return None
    elif length > sys.maxsize:
        print("out of scope")
        return None
    largest, second = max(l), min(l)
    if largest == second:
        return None
    for item in l:
        if item > second and item < largest:
            second = item
    return second

findSecondTest.py 測驗代碼

import unittest
from hyang.python3.interview.boyan.findSecondItem import findSecond
#from findSecondItem import findSecond
 
 
class Test_findSecondItem(unittest.TestCase):
 
    # 如果跑所有用例，只運行一次前提條件和結束條件，則用setupclass()和teardownclass()
    @classmethod
    def setUpClass(cls):
        print("在所有的測驗用例執行之前，只執行一次============")
 
    @classmethod
    def tearDownClass(cls):
        print("在所有的測驗用例執行之后，只執行一次============")
 
    # empty list
    def test_findSecondItem_01(self):
        l1 = list()
        assert (findSecond(l1) == None)
 
    # one item in list
    def test_findSecondItem_02(self):
        l1 = [2]
        assert (findSecond(l1) == None)
 
    # No item in list after remove non-number items
    def test_findSecondItem_03(self):
        l1 = [None, "abc", "xyz"]
        assert (findSecond(l1) == None)
 
    # one item in list after remove non-number items
    def test_findSecondItem_04(self):
        l1 = [None, 3, "abc"]
        assert (findSecond(l1) == None)
 
    # duplated largest number
    def test_findSecondItem_05(self):
        l1 = [32, None, 12, "abc", 8, 6, 36, 3, 32, 4, 36, 9, 25, '35', 36]
        assert (findSecond(l1) == 32)
 
 
# python3中寫不寫都會執行的
if __name__ == '__main__':
    unittest.main()

方法二：findSecondNum.py 從串列去掉所有的最大元素，再在串列中找一個最大就是第二大元素，

import sys
from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems
 
def findSecond(l):
    l = removeInvalidItems(l)
    length = len(l)
    if length == 0:
        print("there is no number item in the list")
        return None
    elif length == 1:
        print("there is only one number item, it's ", l[0])
        return None
    elif length > sys.maxsize:
        print("out of scope")
        return None
    largest=max(l)
    largest_count=l.count(largest)
    while largest_count>0:  #remove all the largest item
        l.remove(largest)
        largest_count-=1
    if len(l)==0:
        return None
    else:
        return max(l)

方法三：利用內置的list.sort，但是要去掉重復元素

import sys
from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems
 
def findSecond(l):
    removeInvalidItems(l)
    l2 = list(set(l))   # remove duplicated items
    l2.sort(reverse=True)
    length=len(l2)
    if length>=2:
        return l2[1]
    else:
        return None

方法四：與方法三類似，利用內置的heapq.nlargest，也需要去掉重復元素

import sys, heapq
from hyang.python3.interview.boyan.findSecondUtil import removeInvalidItems
 
def findSecond(l):
    removeInvalidItems(l)
    l2 = list(set(l))   # remove duplicated items
    length = len(l)
    length = len(l2)
    if length >= 2:
        return heapq.nlargest(2, l2)[1]
    else:
        return None

最后需要注意執行程式所在路徑，見下圖，可以結合自己的配置來調整，

標籤：Python

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