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面试被问傻！5亿个数大文件怎么排序？

曾经被问傻的一道面试题分享给你：

给你1个文件 bigdata ，大小4663M，5亿个数，文件中的数据随机，如下一行一个整数：

现在要对这个文件进行排序，怎么搞？

内部排序

先尝试内排，选2种排序方式：

3路快排：

private final int cutoff = 8;  
  
public <T> void perform(Comparable<T>[] a) {  
  perform(a,0,a.length - 1);  
}  
  
private <T> int median3(Comparable<T>[] a,int x,int y,int z) {  
  if(lessThan(a[x],a[y])) {  
    if(lessThan(a[y],a[z])) {  
      return y;  
    }  
    else if(lessThan(a[x],a[z])) {  
      return z;  
    }else {  
      return x;  
    }  
  }else {  
    if(lessThan(a[z],a[y])){  
      return y;  
    }else if(lessThan(a[z],a[x])) {  
      return z;  
    }else {  
      return x;  
    }  
  }  
}  
  
private <T> void perform(Comparable<T>[] a,int low,int high) {  
  int n = high - low + 1;  
  
  if(n <= cutoff) {  
    InsertionSort insertionSort = SortFactory.createInsertionSort();  
    insertionSort.perform(a,low,high);  
  
  }else if(n <= 100) {  
    int m = median3(a,low,low + (n >>> 1),high);  
    exchange(a,m,low);  
  
  }else {  
    int gap = n >>> 3;  
    int m = low + (n >>> 1);  
    int m1 = median3(a,low,low + gap,low + (gap << 1));  
    int m2 = median3(a,m - gap,m,m + gap);  
    int m3 = median3(a,high - (gap << 1),high - gap,high);  
    int ninther = median3(a,m1,m2,m3);  
    exchange(a,ninther,low);  
  }  
  
  if(high <= low)  
    return;  
  
  int lt = low;  
  
  int gt = high;  
  
  Comparable<T> pivot = a[low];  
  int i = low + 1;  
  
  
  
  while (i <= gt) {  
    if(lessThan(a[i],pivot)) {  
  
      exchange(a,lt++,i++);  
    }else if(lessThan(pivot,a[i])) {  
      exchange(a,i,gt--);  
    }else{  
      i++;  
    }  
  }  
  
  
  perform(a,low,lt - 1);  
  perform(a,gt + 1,high);  
}

归并排序：

private final int cutoff = 8;  
  
  
@Override  
public <T> void perform(Comparable<T>[] a) {  
  Comparable<T>[] b = a.clone();  
  perform(b, a, 0, a.length - 1);  
}  
  
private <T> void perform(Comparable<T>[] src,Comparable<T>[] dest,int low,int high) {  
  if(low >= high)  
    return;  
  
  
  if(high - low <= cutoff) {  
    SortFactory.createInsertionSort().perform(dest,low,high);  
    return;  
  }  
  
  int mid = low + ((high - low) >>> 1);  
  perform(dest,src,low,mid);  
  perform(dest,src,mid + 1,high);  
  
  
  if(lessThanOrEqual(src[mid],src[mid+1])) {  
    System.arraycopy(src,low,dest,low,high - low + 1);  
  }  
  
  
  merge(src,dest,low,mid,high);  
}  
  
private <T> void merge(Comparable<T>[] src,Comparable<T>[] dest,int low,int mid,int high) {  
  
  for(int i = low,v = low,w = mid + 1; i <= high; i++) {  
    if(w > high || v <= mid && lessThanOrEqual(src[v],src[w])) {  
      dest[i] = src[v++];  
    }else {  
      dest[i] = src[w++];  
    }  
  }  
}

数据太多，递归太深 ->栈溢出？加大Xss？

数据太多，数组太长 -> OOM？加大Xmx?

耐心不足，没跑出来.而且要将这么大的文件读入内存，在堆中维护这么大个数据量，还有内排中不断的拷贝，对栈和堆都是很大的压力，不具备通用性。

sort命令来跑

sort -n bigdata -o bigdata.sorted

跑了多久呢？ 24分钟.

为什么这么慢？

粗略的看下我们的资源：

内存

jvm-heap/stack，native-heap/stack,page-cache，block-buffer

数据量很大，函数调用很多，系统调用很多，内核/用户缓冲区拷贝很多，脏页回写很多，io-wait很高，io很繁忙，堆栈数据不断交换至swap，线程切换很多，每个环节的锁也很多.

总之，内存吃紧，问磁盘要空间，脏数据持久化过多导致cache频繁失效，引发大量回写，回写线程高，导致cpu大量时间用于上下文切换，一切，都很糟糕，所以24分钟不细看了，无法忍受.

位图法

private BitSet bits;  
  
public void perform(  
    String largeFileName,  
    int total,  
    String destLargeFileName,  
    Castor<Integer> castor,  
    int readerBufferSize,  
    int writerBufferSize,  
    boolean asc) throws IOException {  
  
  System.out.println("BitmapSort Started.");  
  long start = System.currentTimeMillis();  
  bits = new BitSet(total);  
  InputPart<Integer> largeIn = PartFactory.createCharBufferedInputPart(largeFileName, readerBufferSize);  
  OutputPart<Integer> largeOut = PartFactory.createCharBufferedOutputPart(destLargeFileName, writerBufferSize);  
  largeOut.delete();  
  
  Integer data;  
  int off = 0;  
  try {  
    while (true) {  
      data = largeIn.read();  
      if (data == null)  
        break;  
      int v = data;  
      set(v);  
      off++;  
    }  
    largeIn.close();  
    int size = bits.size();  
    System.out.println(String.format("lines : %d ,bits : %d", off, size));  
  
    if(asc) {  
      for (int i = 0; i < size; i++) {  
        if (get(i)) {  
          largeOut.write(i);  
        }  
      }  
    }else {  
      for (int i = size - 1; i >= 0; i--) {  
        if (get(i)) {  
          largeOut.write(i);  
        }  
      }  
    }  
  
    largeOut.close();  
    long stop = System.currentTimeMillis();  
    long elapsed = stop - start;  
    System.out.println(String.format("BitmapSort Completed.elapsed : %dms",elapsed));  
  }finally {  
    largeIn.close();  
    largeOut.close();  
  }  
}  
  
private void set(int i) {  
  bits.set(i);  
}  
  
private boolean get(int v) {  
  return bits.get(v);  
}

nice!跑了190秒，3分来钟.

以核心内存 4663M/32 大小的空间跑出这么个结果，而且大量时间在用于I/O，不错.问题是，如果这个时候突然内存条坏了1、2根，或者只有极少的内存空间怎么搞？

外部排序