[英]CodeChef TurboSort (Sorting using int vs Integer )
給定數字列表,您將按非遞減順序對它們進行排序。 輸入
t - 列表中的數字數,然后t行跟隨[t <= 10 ^ 6]。 每行包含一個整數:N [0 <= N <= 10 ^ 6]輸出
以非遞減順序輸出給定數字。 例
輸入:5 5 3 6 7 1輸出:1 3 5 6 7
首先使用文字int值並使用Arrays.sort()函數實現,該函數使用Quicksort Algo對文字進行排序(最差情況n ^ 2,平均情況 - nlogn)
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int num = in.nextInt();
int[] n = new int[num];
for (int i = 0; i < n.length; i++) {
n[i] = in.nextInt();
}
Arrays.sort(n);
for (int i = 0; i < n.length; i++) out.println(n[i]);
out.close();
}
}
class InputReader {
private BufferedReader reader;
private StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream));
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
下一個實現是將int文字存儲和排序為Integer對象,並使用Arrays.sort()方法,該方法現在使用MergeSort算法對Integer對象進行排序,以保證nlogn性能
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.io.InputStream;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int T = in.nextInt();
Integer[] ARR = new Integer[T];
for (int i = 0; i < T; i++) ARR[i] = in.nextInt();
Arrays.sort(ARR);
for (int i : ARR) out.println(i);
out.close();
}
}
class InputReader {
private BufferedReader reader;
private StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream));
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
然而現在的問題是,根據邏輯,mergesort算法(即整數對象排序實現)相對於Quicksort算法應該花費更少或相等的時間),即int文字排序實現花費更少的時間......
整數對象排序實現 - 0.94sec int文字排序實現 - 0.53秒
我錯過了什么嗎? 超時的原因是什么? 是因為自動裝箱和自動裝箱?!這就是這個多余時間的原因......
對於初學者來說,實踐中的合並排序和快速排序都具有相似的性能。 事實上,快速排序通常會略微優於隨機數據。 但即使合並排序稍好一些,排序整數總是會明顯變慢,因為排序對象比基元更難。 它們不適用於緩存和原語。
排序需要更長的時間主要是因為使用Integer,您正在存儲一個對象,這是一個很大的開銷。
我想感謝你提醒我,我有一個我已經傾倒了很長時間的codechef帳戶。這是我當時做的解決方案花了我.20秒運行代碼有點大希望你找到這個有用的謝謝..
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintStream;
import java.lang.reflect.Field;
import java.nio.ByteBuffer;
import java.nio.channels.FileChannel;
class Reader
{
private static final int BUFSIZE = 0x10000;
private final byte[] buffer = new byte[BUFSIZE];
private final ByteBuffer bb = ByteBuffer.wrap(buffer);
private final FileChannel channel;
int bufSize = -1; // non empty buffer
int bufOffset = 0; // non valid buffer
private FileInputStream getFileInputStream(InputStream in)
{
try
{
if (in instanceof BufferedInputStream)
{
Field field = in.getClass().getSuperclass().getDeclaredField("in");
field.setAccessible(true);
return (FileInputStream) field.get(in);
}
}
catch (Throwable e)
{
e.printStackTrace();
}
return (FileInputStream) in;
}
Reader(InputStream in) throws IOException
{
this.channel = this.getFileInputStream(in).getChannel();
}
void fetchBuffer() throws IOException
{
bb.clear();
bufSize = channel.read(bb);
bufOffset = 0;
}
boolean isFinished()
{
return bufSize <= 0;
}
private int peek() throws IOException
{
if (bufOffset < bufSize)
return buffer[bufOffset];
fetchBuffer();
if (bufSize > 0)
return buffer[0];
return -1;
}
private void skipSpace() throws IOException
{
int v = peek();
while (v <= ' ' && v != -1)
{
bufOffset++;
v = peek();
}
}
void nextLine() throws IOException
{
int v = peek();
while (v != -1 && v != '\n' && v != '\r')
{
bufOffset++;
v = peek();
}
if (v == '\r')
{
bufOffset++;
v = peek();
if (v == '\n')
bufOffset++;
}
else if (v == '\n')
{
bufOffset++;
v = peek();
if (v == '\r')
bufOffset++;
}
}
int readInt() throws IOException
{
skipSpace();
int result = 0;
int v = peek();
while (v > ' ')
{
result = result * 10 + v - '0';
bufOffset++;
v = peek();
}
return result;
}
}
class Writer
{
private static final int BUFSIZE = 0x10000;
private final FileOutputStream fos;
private final byte[] buffer = new byte[BUFSIZE];
private int offset = 0;
private FileOutputStream getFileOutputStream(PrintStream out)
{
try
{
Field field = out.getClass().getSuperclass().getDeclaredField("out");
field.setAccessible(true);
OutputStream os = (OutputStream) field.get(out);
if (os instanceof BufferedOutputStream)
{
BufferedOutputStream bos = (BufferedOutputStream) os;
field = bos.getClass().getSuperclass().getDeclaredField("out");
field.setAccessible(true);
return (FileOutputStream) field.get(bos);
}
return (FileOutputStream) field.get(out);
}
catch (Throwable e)
{
e.printStackTrace();
}
return null;
}
Writer(PrintStream out) throws IOException
{
fos = getFileOutputStream(out);
}
private static final int[] boundaries = new int[]
{
9, 99, 999, 9999, 99999, 999999, 9999999,
99999999, 999999999
};
private static final int[] divs = new int[]
{
1, 10, 100, 1000, 10000, 100000, 1000000,
10000000, 100000000
};
private static final byte[] numbers = "0123456789".getBytes();
void writeln(int number) throws IOException
{
if (offset > BUFSIZE - 100)
flush();
int index;
for (index = 0; index < boundaries.length; index++)
if (number <= boundaries[index])
break;
for (; index >= 0; index--)
{
int mult = number / divs[index];
buffer[offset++] = numbers[mult];
number -= mult * divs[index];
}
buffer[offset++] = '\n';
}
void flush() throws IOException
{
if (offset > 0)
{
fos.write(buffer, 0, offset);
offset = 0;
}
}
}
class Solution {
public static void main(String[] args) throws java.lang.Exception {
Reader r=new Reader(System.in);
Writer w=new Writer(System.out);
int x,k;
int[] arr2 = new int[1000000];
x = r.readInt();
for (int i = 0; i < x; i++) {
arr2[r.readInt()]++;
}
for (int i = 0; i < 1000000; i++) {
k= arr2[i];
while(k-- > 0){
w.writeln(i);
}
}
w.flush();
}
}
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