题目：原题链接（简单）
标签：字符串
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( 1 ) | 36ms (84.89%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def read(self, buf, n):
now, remain = 0, n
while True:
cache = [" "] * 4
size = read4(cache)
for i in range(min(remain, size)):
buf[now] = cache[i]
now += 1
remain -= size
if size < 4 or remain <= 0:
return now
