归并算法的递归和非递归实现

package Sort;

public class MergeSort {
	public static void main(String[] args) {
		int[] sourceData = new int[]{6,1,3,5,2,8,7,0,9,4,11,10,12};
		int[] workSpace = new int[sourceData.length];
		MergeSort ms = new MergeSort();
//		ms.mergeSort(sourceData, workSpace, 0, workSpace.length - 1);
		ms.mergeSortByLoop(sourceData, workSpace);
		for(int i: sourceData)
			System.out.print(i + " ");
		System.out.println();
	}
	
	/**
	 * 
	 * 递归法
	 */
	public void mergeSort(int[] data, int[] workSpace, int low, int high) {
		if(low < high) {
			int mid = (low + high) / 2;
			mergeSort(data, workSpace, low, mid);
			mergeSort(data, workSpace, mid + 1, high);
			merge(data, workSpace, low, mid, high);
		}
	}
	
	public void merge(int[] data, int[] workSpace, int low, int mid, int high) {
		int index = 0;
		int rightLower = mid + 1;
		int copyLow = low;
		while(low <= mid && rightLower <= high) {
			if(data[low] <= data[rightLower])
				workSpace[index++] = data[low++];
			else
				workSpace[index++] = data[rightLower++];
		}
		while(low <= mid)
			workSpace[index++] = data[low++];
		while(rightLower <= high)
			workSpace[index++] = data[rightLower++];
		
		for(int i = 0; i < index; i++) 
			data[copyLow + i] = workSpace[i];
	}
	
	/**
	 * 非递归法
	 * 设置步长，从1开始，每次增加2的幂
	 * 对指定步长个数进行排序，分两种情况
	 * 第一种：当前步长大小的数组的末尾值小于数组长度，即可以完整合并步长个数的情况
	 * 第二种：当前步长大小的数组的末尾值超出数组大小，则表明不能正好合并，也分两种情况讨论
	 *	 第一：当剩余的值小于或等于步长，排序的mid值的计算方法为先求和在除2
	 *	 第二：当剩余值大于step，下表计算方法为从i起一个完整的步长大小
	 */
	
	public void mergeSortByLoop(int[] data, int[] workSpace) {
		for(int step = 1; step < workSpace.length; step *= 2)
			for(int i = 0; i < workSpace.length; i += 2 * step) {
				if((i + 2 * step - 1) <= workSpace.length - 1)
					merge(data, workSpace, i, i + step - 1, i + 2 * step - 1);
				else {
					if(workSpace.length - 1 - i <= step)
						merge(data, workSpace, i, (i + workSpace.length - 1) / 2, workSpace.length - 1);
					else
						merge(data, workSpace, i, i + step - 1, workSpace.length - 1);
				}
			}
	}
	
}