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摘要：題目鏈接和那道不同的是這次，問題就是當(dāng)前的可能存在多讀了幾個(gè)字節(jié)，那么下一次的時(shí)候要先算上上次多讀的部分，所以要保存上次讀的。和讀一次一樣有兩種要考慮的讀完了沒讀完，但是裝滿了
158. Read N Characters Given Read4 II - Call multiple times
題目鏈接：https://leetcode.com/problems...
和那道read n不同的是這次call multiple times，問題就是當(dāng)前的call可能存在多讀了幾個(gè)字節(jié)，那么下一次call read的時(shí)候要先算上上次多讀的部分，所以要保存上次讀的。和讀一次一樣有兩種要考慮的case：
file讀完了： read4(buf[]) == 0
file沒讀完，但是buf裝滿了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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