归并排序

归并排序利用了分治和递归。归并排序的流程如下：

• 以整个数组为对象执行mergeSort
• mergeSort如下所示：
  1. 将给定的元素分割为两部分
  2. 对这两部分分别执行mergeSort排序
  3. 通过merge操作将两个已排完序的数组合并成一个

要注意的一点是合并操作时，我们新建一个临时数组，然后利用这两个数组已经排序的特点，只需比较((n_1+n_2-1))次。

题目链接： http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_5_B

Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.

Merge(A, left, mid, right)
  n1 = mid - left;
  n2 = right - mid;
  create array L[0...n1], R[0...n2]
  for i = 0 to n1-1
    do L[i] = A[left + i]
  for i = 0 to n2-1
    do R[i] = A[mid + i]
  L[n1] = SENTINEL
  R[n2] = SENTINEL
  i = 0;
  j = 0;
  for k = left to right-1
    if L[i] <= R[j]
      then A[k] = L[i]
           i = i + 1
      else A[k] = R[j]
           j = j + 1

Merge-Sort(A, left, right){
  if left+1 < right
    then mid = (left + right)/2;
         call Merge-Sort(A, left, mid)
         call Merge-Sort(A, mid, right)
         call Merge(A, left, mid, right)

Input

In the first line n is given. In the second line, n integers are given.

Output

In the first line, print the sequence S. Two consequtive elements should be separated by a space character.

In the second line, print the number of comparisons.

Constraints

• n ≤ 500000
• 0 ≤ an element in S ≤ 109

Sample Input 1

10
8 5 9 2 6 3 7 1 10 4

Sample Output 1

1 2 3 4 5 6 7 8 9 10
34

题目中已经给出了伪代码。和前面线性搜索中设置标记简化程序一样，我们在两个数组进行合并时也设置了标记，将数组的末尾放一个无限大的数。

参考代码如下：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class MergeSort {

    static int count = 0;

    // right代表的是最后一个元素的后一个元素
    public static void mergeSort(int[] num, int left, int right){
        if (left+1 < right){
            int mid = (left + right) / 2;
            mergeSort(num, left, mid);
            mergeSort(num, mid, right);
            merge(num, left, mid, right);
        }
    }

    private static void merge(int[] num, int left, int mid, int right) {
        int n1 = mid - left;
        int n2 = right - mid;

        int[] L = new int[n1+1];
        int[] R = new int[n2+1];
        for (int i=0; i<n1; i++){
            L[i] = num[left+i];
        }
        for (int i=0; i<n2; i++){
            R[i] = num[mid+i];
        }
        L[n1] = Integer.MAX_VALUE;
        R[n2] = Integer.MAX_VALUE;

        int i = 0;
        int j = 0;
        for (int k=left; k<right; k++){
            count++;
            if (L[i] <= R[j]){
                num[k] = L[i++];
            }else {
                num[k] = R[j++];
            }
        }
    }

    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] strings = br.readLine().split("\s+");
        int[] num = new int[n];
        for (int i=0; i<n; i++){
            num[i] = Integer.parseInt(strings[i]);
        }
        mergeSort(num, 0, n);

        for (int i=0; i<n; i++){
            if (i > 0){
                System.out.print(" " + num[i]);
            }
            else {
                System.out.print(num[i]);
            }

        }
        System.out.println();
        System.out.println(count);
    }
}

参考文献：《挑战程序设计竞赛-算法和数据结构》