Python 中使用列表理解和排序的第 K 个非重复字符
原文:https://www . geeksforgeeks . org/kth-非重复字符-python-使用-列表-理解-ordereddict/
给定一个字符串和一个数字 k，找到字符串中的第 k 个非重复字符。考虑一个包含大量字符和一个小字符集的大输入字符串。如何只对输入字符串做一次遍历就能找到字符？
示例:
Input : str = geeksforgeeks, k = 3
Output : r
First non-repeating character is f,
second is o and third is r.
Input : str = geeksforgeeks, k = 2
Output : o
Input : str = geeksforgeeks, k = 4
Output : Less than k non-repeating
characters in input.
此问题已有解决方案请参考链接。我们可以使用列表理解和排序在 python 中快速解决这个问题。
# Function to find k'th non repeating character
# in string
from collections import OrderedDict
def kthRepeating(input,k):
# OrderedDict returns a dictionary data
# structure having characters of input
# string as keys in the same order they
# were inserted and 0 as their default value
dict=OrderedDict.fromkeys(input,0)
# now traverse input string to calculate
# frequency of each character
for ch in input:
dict[ch]+=1
# now extract list of all keys whose value
# is 1 from dict Ordered Dictionary
nonRepeatDict = [key for (key,value) in dict.items() if value==1]
# now return (k-1)th character from above list
if len(nonRepeatDict) < k:
return 'Less than k non-repeating characters in input.'
else:
return nonRepeatDict[k-1]
# Driver function
if __name__ == "__main__":
input = "geeksforgeeks"
k = 3
print (kthRepeating(input, k))
输出:
r
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