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LeetCode-158-Read-N-Characters-Given-Read4II-Call-multiple-times.java
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LeetCode-158-Read-N-Characters-Given-Read4II-Call-multiple-times.java
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/*
https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/discuss/49601/What-is-the-difference-between-call-once-and-call-multiple-times
Think that you have 4 chars "a, b, c, d" in the file, and you want to call your function twice like this:
read(buf, 1); // should return 'a'
read(buf, 3); // should return 'b, c, d'
All the 4 chars will be consumed in the first call. So the tricky part of this question is how can you preserve the remaining 'b, c, d' to the second call.
https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/discuss/188293/Google-follow-up-question.-Speed-up-the-copy.
Follow Up:
If there is enough space in "buf", we could try to optimize read4 method to copy the file contents directly to "buf" by providing a pointer of "buf". But this requires optimization of buf4 method.
*/
/**
* The read4 API is defined in the parent class Reader4.
* int read4(char[] buf);
*/
public class Solution extends Reader4 {
// 1.
// char[] temp = new char[4];
// int i = 0; // pointer in temp
// int readCount = 0;
// public int read(char[] buf, int n) {
// int j = 0; // pointer in buf
// if(n <= 0) return 0;
// while(true){
// if(i == 0) readCount = read4(temp);
// // copy data from temp to buf
// while(i < readCount && j < n){
// buf[j++] = temp[i++];
// }
// // if char in temp is used up, set i to 0
// if(i == readCount) i = 0;
// // readCount<4 means file is all read, j==n means we get all we need from file
// if(readCount < 4 || j == n) break;
// }
// return j;
// }
// 2.
/**
* @param buf Destination buffer
* @param n Number of characters to read
* @return The number of actual characters read
*/
private int bufCount = 0; // the count of chars in buf4
private int bufPointer = 0; // the pointer in buf4
private char[] buf4 = new char[4];
public int read(char[] buf, int n) {
int total = 0;
while (total < n) {
if (bufPointer == 0) bufCount = read4(buf4);
while (total < n && bufPointer < bufCount) buf[total++] = buf4[bufPointer++];
if (bufCount < 4) break; // there is nothing more to read in read4 method, we reached EOF
if (bufPointer == bufCount) bufPointer = 0; // reset the bufPointer
}
return total;
}
}