158. Read N Characters Given Read4 II - Call multi

SillyMonkey 发布于2019-08-19 10:45 / 2541人阅读

摘要：题目链接和那道不同的是这次，问题就是当前的可能存在多读了几个字节，那么下一次的时候要先算上上次多读的部分，所以要保存上次读的。和读一次一样有两种要考虑的读完了没读完，但是装满了

158. Read N Characters Given Read4 II - Call multiple times

题目链接：https://leetcode.com/problems...

和那道read n不同的是这次call multiple times，问题就是当前的call可能存在多读了几个字节，那么下一次call read的时候要先算上上次多读的部分，所以要保存上次读的。和读一次一样有两种要考虑的case：

file读完了： read4(buf[]) == 0

file没读完，但是buf装满了read4(buf[]) > n - res

public class Solution extends Reader4 {
    /**
     * @param buf Destination buffer
     * @param n   Maximum number of characters to read
     * @return    The number of characters read
     */
    char[] buff = new char[4];
    int start = 0;
    int num = 0;
    public int read(char[] buf, int n) {
        int res = 0;
        while(res < n) {
            // no previous buff remain
            if(start == 0) num = read4(buff);
            // finish reading
            if(num == 0) break;
            // count the remain char to use for next call: start is the next start
            while(res < n && start < num) buf[res++] = buff[start++];
            // clear start if read all: n - res >= num - start
            if(start == num) start = 0;
        }
        return res;
    }
}

157. Read N Characters Given Read4

    public int read(char[] buf, int n) {
        int res = 0;
        char[] temp = new char[4];
        while(res < n) {
            int cur = read4(temp);
            if(cur == 0) break;
            
            int num = Math.min(cur, n - res);
            for(int j = 0; j < num; j++) buf[res++] = temp[j];
        }
        return res;
    }

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