拼多多2019.09.01秋招笔试 算法工程师

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def pdd_1():
 # 根据取模运算，将奇数和偶数分开，分别排序，之后拼接输出
 
 # nums, N = input().split(';')
 
 aa = '555503,772867,756893,339138,399447,40662,859037,628085,855723,974471,599631,636153,581541,174761,948135,411485,554033,858627,402833,546467,574367,360461,566480,755523,222921,164287,420256,40043,977167,543295,944841,917125,331763,188173,353275,175757,950417,284578,617621,546561,913416,258741,260569,630583,252845;10'
 nums, N = aa.split(';')
 nums = map(int, nums.split(','))
 N = int(N)
 
 nums_0, nums_1 = [], []
 for i in nums:
 if i % 2 == 0:
 nums_0.append(i)
 nums_0.sort(reverse=True)
 else:
 nums_1.append(i)
 nums_1.sort(reverse=True)
 
 nums = nums_0 + nums_1
 for i in range(N - 1):
 print(nums[i], end=',')
 print(nums[N], end='')

def pdd_2():
 # DFS 求解，着重找到递归结束的条件
 # DFS 注意顺序，以满足输出的字典序要求，避免二次排序
 
 def check(nums0, new_0, nums1, new_res):
 if not nums0:
 if len(new_0) == len(nums1):
 for i in range(len(new_0)):
 if new_0[i] != nums1[i]:
 return
 res.append(new_res)
 return
 
 num = nums0[0]
 nums0 = nums0[1:]
 
 check(nums0, new_0, nums1, new_res + ['d'])
 check(nums0, [num] + new_0, nums1, new_res + ['l'])
 check(nums0, new_0 + [num], nums1, new_res + ['r'])
 
 # N = int(input())
 N = 1
 for i in range(N):
 # nums_0 = list(map(int, input().split()))
 # nums_1 = list(map(int, input().split()))
 
 nums_0 = [1, 2, 3]
 nums_1 = [3]
 print('{')
 if set(nums_0) > set(nums_1):
 res = []
 check(nums_0, [], nums_1, [])
 for j in res:
 print(' '.join(j))
 print('} ')