算法导论Java实现-选择排序（习题2.2-2）

 

package lhz.algorithm.chapter.two;

/**
 * 《选择排序》，《算法导论》习题2.2-2
 * Consider sorting n numbers stored in array A by first
 * finding the smallest element of A and exchanging it with the element in A[1].
 * Then find the second smallest element of A, and exchange it with A[2].
 * Continue in this manner for the first n - 1 elements of A. Write pseudocode
 * for this algorithm, which is known as selection sort . What loop invariant
 * does this algorithm maintain? Why does it need to run for only the first n -
 * 1 elements, rather than for all n elements? Give the best-case and worst-
 * case running times of selection sort in Θ- notation.
 * 伪代码：
 *  for i <- 1 to length[A]-1
 *      key <- A[i]
 *      index <- i
 *      for j <- 2 to length[A]
 *          key = key < A[j] ? key : A[j];
            index = key < A[j] ? index : j;
 *      A[index] = A[i];
        A[i] = key;
 * @author lihzh(苦逼 coder)
 * 本文地址：http://mushiqianmeng.blog.51cto.com/3970029/728872
 */
public class SelectionSort {

    private static int[] input = new int[] { 2, 1, 5, 4, 9, 8, 6, 7, 10, 3 };

    /**
     * @param args
     */
    public static void main(String[] args) {
        for (int i=0; i<input.length-1; i++) {//复杂度数量级：n
            int key = input[i];
            int index = i;
            //比较当前值和下一个值的关系，记录下较小值的值和索引数，用于交换。
            for (int j=i+1; j<input.length; j++) {//复杂度：1+2+...+(n-1)=Θ(n^2)
                key = key < input[j] ? key : input[j];
                index = key < input[j] ? index : j;
            }
            input[index] = input[i];
            input[i] = key;
        }
        /*
         * 复杂度分析：
         * 最坏情况下，复杂度为：n*n=n^2(若略微精确的计算即为：n-1+1+2+...+n-1=(2+n)*(n-1)/2,
         * 所以复杂度仍为n^2。
         * 最优情况下，由于不论原数组是否排序好，均需要全部遍历以确定当前的最小值，所以复杂度不变仍未n^2。
         *
         */
        //打印数组
        printArray();
    }

    private static void printArray() {
        for (int i : input) {
            System.out.print(i + " ");
        }
    }

}