标签:for elf getch 次数 main self 它的 turn char
给定一个字符串,找到它的第一个不重复的字符,并返回它的索引。如果不存在,则返回 -1。
s = "leetcode" 返回 0. s = "loveleetcode", 返回 2.
class solution:
def getChar(self,s):
setS = set(s)
dict = {}
for i in setS:
dict[i] = s.count(i) #该方法返回子字符串在字符串中出现的次数。
for index,value in enumerate(s):
print(‘j‘+str(index)+‘ , k‘+str(value))
if dict.get(value) == 1:
return index
return -1
def getChar2(self,s):
return min([s.find(c) for c in ‘abcdefghijklmnopqrstuvwxyz‘ if s.count(c)==1] or [-1])
if __name__ == "__main__":
solution = solution()
index = solution.getChar(‘leetcode‘)
index2 = solution.getChar(‘loveleetcode‘)
print(index)
print(index2)
print(solution.getChar2(‘leetcode‘))
print(solution.getChar2(‘loveleetcode‘))
标签:for elf getch 次数 main self 它的 turn char
原文地址:https://www.cnblogs.com/turningli/p/12446062.html