归并排序详解与比较分析-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_36994568/article/details/122352774

本文深入探讨了归并排序算法，包括不使用警戒值的传统MergeSort、三路归并排序及其优化。通过详细步骤阐述了如何进行分段和元素选择，并对算法进行了证明，分析了比较次数，为理解和实现归并排序提供了全面指导。

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1. 证明 $f(N)=NlgN+O(N)$ 与 $f(N)=\Theta(NlogN)$ 相等。

2. 不使用警戒值的MergeSort

//文件名：MergeSort.java
//编译：javac MergeSort.java
//执行：java MergeSort -ea  #-ea: enable assert

import java.util.Random;

public class MergeSort {
    public static int[] b = new int[65535];
    public static int[] c = new int[65535];

    public static void sort(int[] a, int lo, int hi) {
        if (hi <= lo) return;
        int mid = lo + (hi - lo) / 2;
        MergeSort.sort(a, lo, mid);
        MergeSort.sort(a, mid + 1, hi);

        for (int k = lo; k <= mid; k++) b[k - lo] = a[k];
        for (int k = mid + 1; k <= hi; k++) c[k - mid - 1] = a[k];

        int i = 0, j = 0;
        for (int k = lo; k <= hi; k++) {
            if (!(i < mid - lo + 1)) {
                a[k] = c[j++];
            } else if (!(j < hi - mid)) {
                a[k] = b[i++];
            } else {
                if (c[j] < b[i]) a[k] = c[j++];
                else a[k] = b[i++];
            }
        }
    }

    public static void main(String[] args) throws Exception {
        int size = 65535;
        int[] a = new int[size];
        Random random = new Random();
        for (int i = 0; i < size; i++) a[i] = random.nextInt(65535 * 2);
        MergeSort.sort(a, 0, size - 1);

        for (int i = 1; i < size; i++) assert a[i - 1] <= a[i];
    }
}

3. 三路归并排序（三向MergeSort），使用警戒值。

I.分段 $[lo, hi]$ ，假设共 $N$ 个元素，分为三段： $\lfloor \frac N3 \rfloor,N - \lfloor \frac N3 \rfloor - \lceil \frac N3 \rceil , \lceil \frac N3 \rceil$

II.每一次选择元素保证使用两次比较

//文件名：TriMergeSort.java
//编译：javac TriMergeSort.java
//执行：java TriMergeSort -ea # -ea: enable assert
import java.util.Random;

public class TriMergeSort {
    public static int[] b = new int[65535];
    public static int[] c = new int[65535];
    public static int[] d = new int[65535];

    public static void sort(int[] a, int lo, int hi) {
        if (hi <= lo) return;

        // divide [lo, hi] into [lo,barrier_1], [barrier_1+1, barrier_2-1], [barrier_2, hi]
        int barrier_1 = lo + (hi - lo + 1) / 3 - 1;
        int barrier_2 = hi - (hi - lo + 3) / 3 + 1;

        TriMergeSort.sort(a, lo, barrier_1);
        TriMergeSort.sort(a, barrier_1 + 1, barrier_2 - 1);
        TriMergeSort.sort(a, barrier_2, hi);

        for (int k = lo; k <= barrier_1; k++) b[k - lo] = a[k];
        for (int k = barrier_1 + 1; k <= barrier_2 - 1; k++) c[k - barrier_1 - 1] = a[k];
        for (int k = barrier_2; k <= hi; k++) d[k - barrier_2] = a[k];

        b[barrier_1 - lo + 1] = Integer.MAX_VALUE;
        c[barrier_2 - barrier_1 - 1] = Integer.MAX_VALUE;
        d[hi - barrier_2 + 1] = Integer.MAX_VALUE;

        int i = 0, j = 0, k = 0;
        for (int m = lo; m <= hi; m++) {
            if (b[i] <= c[j]) {
                if (d[k] <= b[i]) a[m] = d[k++];
                else a[m] = b[i++];
            } else {
                if (d[k] <= c[j]) a[m] = d[k++];
                else a[m] = c[j++];
            }
        }
    }

    public static void main(String[] args) throws Exception {
        int size = 65535;
        int[] a = new int[size];
        Random random = new Random();
        for (int i = 0; i < size; i++) a[i] = random.nextInt(65535 * 2);
        TriMergeSort.sort(a, 0, size - 1);

        for (int i = 1; i < size; i++) assert a[i - 1] <= a[i];
    }
}

4. MergeSort证明 $C_N=\sum_{1\le k <N}(\lfloor lgk \rfloor + 2)$