#初學算法# 歸并排序與冒泡排序算法對比

        最近剛看算法導論，想自己比較一下冒泡排序跟歸并排序。

        歸并排序的算法複雜度：nlogn

        冒泡排序的算法複雜度：n^2

        第一次發部落格，各位大蝦有什麼建議盡管提，非常感謝！

public class Sort {

	/**
	 * @param 
	 * @author darkhorse_pxf
	 */
	static int count=1;  //用于計算歸并排序操作次數
	static int count2=1;  用于計算冒泡排序操作次數
	public static void main(String[] args) {
		int a[]={54,47,15,475,21,57,67,18,88,245};
		int b[]={54,47,15,475,21,57,67,18,88,245};
		System.out.println("歸并排序:");
		mergeSort(a,0,a.length-1);   //歸并排序
		System.out.println("歸并排序共執行"+count+"次");
		System.out.println("冒泡排序:");
		bubbleSort(b);      //冒泡排序
		System.out.println("冒泡排序共執行"+count2+"次");
		/**測試Merge
		int a[]={1,3,5,7,9,2,4,6,8,10};
		
		Merge.Merge(a, 0, 4, a.length-1);
		for(int i=0;i<a.length;i++){
			System.out.print(a[i]+"   ");
		}
		System.out.println();
		*/
	}
	private static void bubbleSort(int[] b) {   //冒泡排序
		int k;
		for(int i=0;i<b.length;i++){
			
			for(int j=i;j<b.length;j++){
				if(b[i]<=b[j]){
					k=b[i];
					b[i]=b[j];
					b[j]=k;
					count2++;
				}
			}
			for(int x=0;x<b.length;x++){
				System.out.print(b[x]+"   ");
			}
			System.out.println();
		}
	}
	public static void mergeSort(int[] numbers,int firstNumber,int lastNumber) {
		if(firstNumber<lastNumber){
			int midleNumber=(firstNumber+lastNumber)/2;
			mergeSort(numbers,firstNumber,midleNumber);
			mergeSort(numbers,midleNumber+1,lastNumber);
			Merge(numbers, firstNumber, midleNumber, lastNumber);
			for(int i=0;i<numbers.length;i++){
				System.out.print(numbers[i]+"   ");
			}
			System.out.println();
			
		}
	}
	public static void Merge(int numbers[],int p,int q,int r){
		int n1=q-p+1;
		int n2=r-q;
		int Left[]=new int[n1];
		int Right[]=new int[n2];
		for(int i=0;i<Left.length;i++){
			Left[i]=numbers[i+p];
		}
		for(int j=0;j<Right.length;j++){
			Right[j]=numbers[q+1+j];
		}
		int i=0;
		int j=0;
		while(p<=r){
			if(i==Left.length){
				while(j<Right.length){
					numbers[p++]=Right[j++];
				}
				break;
			}
			else if(j==Right.length){
				while(i<Left.length){
					numbers[p++]=Left[i++];
				}
				break;
			}
			if(Left[i]<=Right[j]){
				numbers[p++]=Left[i++];
			}else{
				numbers[p++]=Right[j++];
			}
			count++;
		}
		
	}

}
           

運作結果：