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面试过程中, 算法几乎变成了一个逃不过去的坎儿, 而在算法的面试中, 关于字符串的问题很常见,也很基础,今天我们就来讲几个非常有代表意义的面试题。
def count_s(s):
if len(s)<=0 or not isinstance(s, str):
return
base = s[0]
result = []
count = 1
for i in s[1:]:
if i == base:
count +=1
else:
result.append(base+str(count))
count =1
base = i
result.append(base + str(count))
return result
print(count_s('abbcccdxt'))
def count_s(s):
if len(s)<=0 or not isinstance(s, str):
return
base = s[0]
result = []
count = 1
for i in s[1:]:
if i == base:
count +=1
else:
result.append(base*count)
count =1
base = i
result.append(base *count)
return result
print(count_s('abbcccdxt'))
def count_s(s):
if len(s)<=0 or not isinstance(s, str):
return
base = s[0]
result = []
count = 1
for i in s[1:]:
if i == base:
count +=1
else:
if count ==3:
result.append(base*count)
count =1
base = i
if count == 3:
result.append(base * count)
return result
print(count_s('abbcccdxt'))
from collections import Counter
def count_s(s):
if len(s)<=0 or not isinstance(s, str):
return
return [s[number:number+3] for number in range(len(s)-2)]
print(Counter(count_s('abccbbcc')))