歸併算法的遞歸和非遞歸實現

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();
	}
	
	/**
	 * 
	 * 遞歸法
	 */
	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];
	}
	
	/**
	 * 非遞歸法
	 * 設置步長，從1開始，每次增加2的冪
	 * 對指定步長個數進行排序，分兩種情況
	 * 第一種：當前步長大小的數組的末尾值小於數組長度，即可以完整合並步長個數的情況
	 * 第二種：當前步長大小的數組的末尾值超出數組大小，則表明不能正好合並，也分兩種情況討論
	 *	 第一：當剩餘的值小於或等於步長，排序的mid值的計算方法爲先求和在除2
	 *	 第二：當剩餘值大於step，下表計算方法爲從i起一個完整的步長大小
	 */
	
	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);
				}
			}
	}
	
}