选择排序的递归和迭代实现(范型)

package edu.cumt.jnotnull;

public class SortArray {
	public static <T extends Comparable<? super T>> void selectionSort(T[] a,
			int n) {
		for (int index = 0; index < n - 1; index++) {
			int indexOfNextSmallest = getIndexOfSmallest(a, index, n - 1);
			swap(a, index, indexOfNextSmallest);
		}
	}
	
	public static <T extends Comparable<? super T>> void selectionSort(T[] a,
			int first, int last) {
		if (first <= last) {
			int indexOfNextSmallest = getIndexOfSmallest(a, first, last);
			swap(a, first, indexOfNextSmallest);
			selectionSort(a, first + 1, last);
		}
	}

	private static <T extends Comparable<? super T>> int getIndexOfSmallest(
			T[] a, int first, int last) {
		T min = a[first];
		int indexOfMin = first;
		for (int index = first + 1; index < last; index++) {
			if (a[index].compareTo(min) < 0) {
				min = a[index];
				indexOfMin = index;
			}
		}
		return indexOfMin;
	}

	private static void swap(Object[] a, int i, int j) {
		Object temp = a[i];
		a[i] = a[j];
		a[j] = temp;
	}

}

 

迭代方法的算法效率

selectionSort执行了n-1次,所以各调用indexofsmallest和swap各n-1次.

在对indexofsmallest的调用中,last是n-1 first是n-2.每一次调用indexofsmallest,琪循环都执行了last-first次.因此这个循环共执行了(n-1)+(n-2)+...+1    也就是n(n-1)/2 .同时由于循环中的操作是O(1),所以选择排序效率是O(n^2).

同时注意,这里讨论与数组中元素的性质无关,可以是有序的 也可以是无序的.但是都是O(n^2).