使用列表理解和OrderedDict在Python中的第K个非重复字符
在本文中，我们将使用ListComprehension和OrderedDict了解Python中的第K个非重复字符。为此，我们借助Python中可用的内置构造。
算法
1. First, we form a dictionary data from the input. 2. Now we count the frequency of each character. 3. Now we extract the list of all keys whose value equals 1. 4. Finally, we return k-1 character.
示例
from collections import OrderedDict
import itertools
def kthRepeating(inp,k):
# returns a dictionary data
dict=OrderedDict.fromkeys(inp,0)
# frequency of each character
for ch in inp:
dict[ch]+=1
# now extract list of all keys whose value is 1
nonRepeatDict = [key for (key,value) in dict.items() if value==1]
# returns (k-1)th character
if len(nonRepeatDict) < k:
return 'no ouput.'
else:
return nonRepeatDict[k-1]
# Driver function
if __name__ == "__main__":
inp = "nhooo"
k = 3
print (kthRepeating(inp, k))输出结果
a
结论
在本文中，我们使用ListComprehension和OrderedDict在Python中找到了第K个非重复字符。