[leetcode]Read N Characters Given Read4 II - Call multiple times
Posted 阿牧遥
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[leetcode]Read N Characters Given Read4 II - Call multiple times相关的知识,希望对你有一定的参考价值。
调用多次readN,所以要重复使用internal buffer
""" The read4 API is already defined for you. @param buf, a list of characters @return an integer def read4(buf): # Below is an example of how the read4 API can be called. file = File("abcdefghijk") # File is "abcdefghijk", initially file pointer (fp) points to ‘a‘ buf = [‘ ‘] * 4 # Create buffer with enough space to store characters read4(buf) # read4 returns 4. Now buf = [‘a‘,‘b‘,‘c‘,‘d‘], fp points to ‘e‘ read4(buf) # read4 returns 4. Now buf = [‘e‘,‘f‘,‘g‘,‘h‘], fp points to ‘i‘ read4(buf) # read4 returns 3. Now buf = [‘i‘,‘j‘,‘k‘,...], fp points to end of file """ class Solution: def __init__(self): self.buf4 = [‘ ‘] * 4 self.buf4Start = 0 self.buf4End = 0 self.isEof = False def read(self, buf, n): """ :type buf: Destination buffer (List[str]) :type n: Number of characters to read (int) :rtype: The number of actual characters read (int) """ cnt = 0 while cnt < n: if self.isEof and self.buf4Start >= self.buf4End: # nothing to read break elif self.buf4Start < self.buf4End: # copy from buf4 buf[cnt] = self.buf4[self.buf4Start] cnt += 1 self.buf4Start += 1 else: # no more in buf4, read4 self.buf4End = read4(self.buf4) self.buf4Start = 0 if self.buf4End == 0: self.isEof = True return cnt
以上是关于[leetcode]Read N Characters Given Read4 II - Call multiple times的主要内容,如果未能解决你的问题,请参考以下文章
LeetCode 158: Read N Characters Given Read4 II
leetcode 157. Read N Characters Given Read4 利用read4实现read --------- java
[leetcode]158. Read N Characters Given Read4 II - Call multiple times 用Read4读取N个字符2 - 调用多次
Leetcode 158: Read N Characters Given Read4 II - Call multiple times
[leetcode]Read N Characters Given Read4 II - Call multiple times