代码如下:
from collections import defaultdict
class Solution:
def numberOfPairs(self, nums: List[int]) -> List[int]:
a=defaultdict(int)
for i in nums:
a[i]+=1
print(a.keys(),a.values(),a)
for i in a.keys():
a[i]%=2
print(Counter(a.values()))
print(Counter(a.values())[0])
return (sum(a.values()))
def stringToIntegerList(input):
return json.loads(input)
def integerListToString(nums, len_of_list=None):
if not len_of_list:
len_of_list = len(nums)
return json.dumps(nums[:len_of_list])
def main():
import sys
import io
def readlines():
for line in io.TextIOWrapper(sys.stdin.buffer, encoding='utf-8'):
yield line.strip('\n')
lines = readlines()
while True:
try:
line = next(lines)
nums = stringToIntegerList(line);
ret = Solution().numberOfPairs(nums)
out = integerListToString(ret);
print(out)
except StopIteration:
break
if __name__ == '__main__':
main()
利用的是defaultdict以及counter类来实现两个统计的,一个是为了统计最初的个数,另一个是为了统计配对后0的个数,之后返回的是个列表,给它俩放一块返回就行。最开始的时候,想着是用零的个数来计算其配对的个数,结果发现不行,因为有的配对完之后不是零,有的配对了好几次,所以要换个方法,建立两个词典,一个记录余数,另一个记录原本的商数,这样的坏处在于浪费的内存比较多,但是还行,能用:
from collections import defaultdict
class Solution:
def numberOfPairs(self, nums: List[int]) -> List[int]:
a=defaultdict(int)
p=defaultdict(int)
for i in nums:
a[i]+=1
p[i]+=1
print(a.keys(),a.values(),a)
for i in a.keys():
a[i]%=2
p[i]//=2
print(sum(p.values()))
print(sum(a.values()))
def stringToIntegerList(input):
return json.loads(input)
def integerListToString(nums, len_of_list=None):
if not len_of_list:
len_of_list = len(nums)
return json.dumps(nums[:len_of_list])
def main():
import sys
import io
def readlines():
for line in io.TextIOWrapper(sys.stdin.buffer, encoding='utf-8'):
yield line.strip('\n')
lines = readlines()
while True:
try:
line = next(lines)
nums = stringToIntegerList(line);
ret = Solution().numberOfPairs(nums)
out = integerListToString(ret);
print(out)
except StopIteration:
break
if __name__ == '__main__':
main()
双百,吓一跳哈哈哈。
力扣上的代码是:
from collections import Counter,defaultdict
class Solution:
def numberOfPairs(self, nums: List[int]) -> List[int]:
a=defaultdict(int)
p=defaultdict(int)
for i in nums:
a[i]+=1
p[i]+=1
print(a.keys(),a.values(),a)
for i in a.keys():
a[i]%=2
p[i]//=2
return [(sum(p.values())),(sum(a.values()))]