小白笔记---------------------自然合并排序算法（感谢bob的指导）

本文介绍了一个名为'自然排序'的算法实现，该算法通过将输入的整数序列按其自然出现的顺序进行分组和排序，实现了较为高效的排序效果。文章详细展示了算法的工作流程，包括读取输入、获取分组计数、排序过程等关键步骤。

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import java.awt.List;
import java.util.LinkedList;
import java.util.Scanner;

public class Naturesort {
	



	public static void main(String[] args) {
		int[] a = read();
		print(a);
		int[] b = getCount(a);
		sort(a,b);
		print(a);
	}
	
	public static int[]  read(){
		LinkedList<Integer> arr = new LinkedList<Integer>();
		Scanner s = new Scanner(System.in);
		int num;
		while((num = s.nextInt()) != 0){
			 arr.add(num); 
		}
		Integer[] n = arr.toArray(new Integer[0]);
		int[] a = new int[n.length];
		for(int i=0; i < n.length; i ++)
		{
			a[i] = n[i].intValue();
		}
		return a;
	}
	public static void print(int[] n){
		for(int i =0; i < n.length ; i++){
			System.out.print(n[i]+" ");
		}
		System.out.println();
	}
	public static int[] getCount(int[] n){
		int i = 0,j = -1,count = 0;
		int[] b = new int[n.length+1];//b数组用来记录分成数组的大小，下表标值+1用来表示数组的个数
		while(i < (n.length -1)){
			if(n[i] <= n[i+1]){
				i++;
			}else{
				b[count] = i - j;
				count++;
				j = i;
				i++;
			}
		}
		b[count++] = n.length - j - 1;
		if(count == 1){
			System.out.println("排序结束！");
			print(n);
			System.exit(0);
		}
		print(b);
		return b;
	}
	


	public static void sort(int[] n,int[] b){
		int k = 0  ,count = 0;
		int j =0,p = 0, q = 0;
		int indexC = 0, indexD = 0;
		while(b[k++] != 0){
			count++;
		}
		for(int i = count/2;i >= 1;i--){
			int[] c = new int[b[j]];
			int[] d = new int[b[j+1]];
			indexC = 0;
			indexD = 0;//注意下标置零，哼重要！
			for(;indexC < c.length ;indexC++){
				c[indexC] = n[p++];
			}
			for(;indexD < d.length ;indexD++){
				d[indexD] = n[p++];
			}
			indexC = 0;
			indexD = 0;//注意下标置零，很重要！！
			
			while(indexC < c.length && indexD < d.length){
				if(c[indexC] <= d[indexD]){
					n[q++] = c[indexC++];
				}else{
					n[q++] = d[indexD++];
				}
			}
			if(indexC < c.length)
			{
				for(;indexC < c.length ; indexC++){
					n[q++] = c[indexC];
				}
			}else if(indexD < d.length){
				for(;indexD < d.length;indexD++){
					n[q++] = d[indexD];
				}
			}
			j = j+ 2; 
		}
		print(n);
		b = getCount(n);
		sort(n,b);
	}
}