zoukankan      html  css  js  c++  java
  • [LeetCode#157] Read N Characters Given Read4

    Problem:

    The API: int read4(char *buf) reads 4 characters at a time from a file.

    The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

    By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.

    Note:
    The read function will only be called once for each test case.

    Analysis:

    This problem is not hard, but it is easy to be wrong. 
    General Idea:
    Since read4(char[] buf) would always filled buf with four characters, no matter how many characters left in the file.
    case: read4(char[] buf) may fill buf with "['a', 'b', x00, x00]", if there are only two characters left in the file.
    
    Thus we could not directly use "read4" over "char[] buf", and we should take advantage of a temp_buffer for this purpose. Thus we could base on the return int of "read 4" to add the character into buf. 
    
    There are possible two situation of the end:
    1. there are not enough characters left in the file. (for the target n) 
    2. n was meeted. 
    
    For this condition, we should maintain a count of copied words.
    if (cur_len < 4 || count == n)
        break;
    
    When we copy the characters from the temp_buffer, we should only copy the valid range.
    1. the actual words we have read (may less than 4)
    2. iff we have already reach the target n. (may just need part of the characters)
    read_len = read4(temp_buffer);
    cur_len = Math.min(read_len, n - count);
    
    Then we should copy those characters into buf. 
    for (int i = 0; i < cur_len; i++)
        buf[count+i] = temp_buffer[i];
    Note: the cur_len's computation is very important along the process!!!

    Solution:

    public class Solution extends Reader4 {
        /**
         * @param buf Destination buffer
         * @param n   Maximum number of characters to read
         * @return    The number of characters read
         */
        public int read(char[] buf, int n) {
            if (buf == null)
                throw new IllegalArgumentException("buf is null");
            if (n <= 0)
                return 0;
            int count = 0, read_len = 0, cur_len = 0;
            char[] temp_buffer = new char[4];
            while (true) {
                read_len = read4(temp_buffer);
                cur_len = Math.min(read_len, n - count);
                for (int i = 0; i < cur_len; i++)
                    buf[count+i] = temp_buffer[i];
                count += cur_len;
                if (cur_len < 4 || count == n)
                    break;
            }
            return count;
        }
    }
  • 相关阅读:
    mysql 使用 insert ignore into和unique实现不插入重复数据功能
    mysql 判断指定条件数据存不存在,不存在则插入
    Unity3D之如何将包大小减少到极致
    Unity3D–Texture图片空间和内存占用分析
    使用Unity3D的50个技巧:Unity3D最佳实践!
    Unity中的Path对应各平台中的Path
    unity 在移动平台中,文件操作路径详解
    unity Mathf 数学运算汇总
    解决ngui在3d场景中 点透的情况
    【整理】unity3d优化总结篇
  • 原文地址:https://www.cnblogs.com/airwindow/p/4811922.html
Copyright © 2011-2022 走看看