Arrays.sort如何實現(xiàn)降序排序

更新時間：2022年11月25日 17:10:33 作者：北冥SP

這篇文章主要介紹了Arrays.sort如何實現(xiàn)降序排序問題，具有很好的參考價值，希望對大家有所幫助。如有錯誤或未考慮完全的地方，望不吝賜教

Arrays.sort實現(xiàn)降序排序

在調(diào)用Arrays.sort()對數(shù)組進行排序時，默認是升序排序的，如果想讓數(shù)組降序排序，有下面兩種方法：

1.Collections的reverseOrder

import java.util.*;
?
public class Main {
? ? public static void main(String[] args) {
// ? ? ? ?注意這里是Integer，不是int
? ? ? ? Integer[] arr={9,8,7,6,5,4,3,2,1};
? ? ? ? Arrays.sort(arr,Collections.reverseOrder());
? ? ? ? for(int i:arr){
? ? ? ? ? ? System.out.println(i);
? ? ? ? }
? ? }
}

2.利用Comparator接口復(fù)寫compare

import java.util.*;
?
public class Main {
? ? public static void main(String[] args) {
? ? ? ? Integer[] arr={9,8,7,6,5,4,3,2,1};
? ? ? ? Comparator cmp=new CMP();
? ? ? ? Arrays.sort(arr,cmp);
? ? ? ? for(int i:arr){
? ? ? ? ? ? System.out.println(i);
? ? ? ? }
? ? }
}
class CMP implements Comparator<Integer>{
? ? @Override //可以去掉。作用是檢查下面的方法名是不是父類中所有的
? ? public int compare(Integer a,Integer b){
// ? ? ? ?兩種都可以，升序排序的話反過來就行
// ? ? ? ?return a-b<0?1:-1;
? ? ? ? return b-a;
? ? }
}

注意：如果需要改變默認的排列方式，不能使用基本類型（int，char等）定義變量，而應(yīng)該用對應(yīng)的類

Arrays.sort底層原理

概述

Collections.sort()方法底層調(diào)用的也是Arrays.sort()方法,下面我們通過測試用例debug,探究一下其源碼,首先說一下結(jié)果,使用到了插入排序，雙軸快排，歸并排序

雙軸快排(DualPivotQuicksort): 顧名思義有兩個軸元素pivot1，pivot2，且pivot ≤
pivot2，將序列分成三段：x < pivot1、pivot1 ≤ x ≤ pivot2、x >pivot2，然后分別對三段進行遞歸。這個算法通常會比傳統(tǒng)的快排效率更高，也因此被作為Arrays.java中給基本類型的數(shù)據(jù)排序的具體實現(xiàn)。

大致流程:

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快速排序部分展開

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案例

	public static void main(String[] args) {
        int[] nums = new int[]{6,5,4,3,2,1};
        List<Integer> list = Arrays.asList(6, 5, 4, 3, 2, 1);
        Arrays.sort(nums);
        Collections.sort(list);
        System.out.println(Arrays.toString(nums));
        System.out.println(list);

    }

運行結(jié)果

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1 進入Arrays.sort()方法

/**
     * Sorts the specified array into ascending numerical order.
     *
     * <p>Implementation note: The sorting algorithm is a Dual-Pivot Quicksort
     * by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm
     * offers O(n log(n)) performance on many data sets that cause other
     * quicksorts to degrade to quadratic performance, and is typically
     * faster than traditional (one-pivot) Quicksort implementations.
     *
     * @param a the array to be sorted
     */
    public static void sort(int[] a) {
        DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);
    }

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方法上的注釋

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2 進入DualPivotQuicksort類內(nèi)部的靜態(tài)方法sort

方法上的注釋

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3 走sort的流程

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1. 排序范圍小于286的數(shù)組使用快速排序

 	// Use Quicksort on small arrays
    if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
    }
    // Merge sort
    ......

2. 進入sort方法,判斷數(shù)組長度是否小于47，小于則直接采用插入排序,否則執(zhí)行3。

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	 // Use insertion sort on tiny arrays
    if (length < INSERTION_SORT_THRESHOLD) {
	   // Insertion sort
	   ......
    }

3. 用公式length/8+length/64+1近似計算出數(shù)組長度的1/7。

// Inexpensive approximation of length / 7
int seventh = (length >> 3) + (length >> 6) + 1;

4. 取5個根據(jù)經(jīng)驗得出的等距點。

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		/*
         * Sort five evenly spaced elements around (and including) the
         * center element in the range. These elements will be used for
         * pivot selection as described below. The choice for spacing
         * these elements was empirically determined to work well on
         * a wide variety of inputs.
         */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

5.將這5個元素進行插入排序

		// Sort these elements using insertion sort
        if (a[e2] < a[e1]) { long t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { long t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { long t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { long t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

6. 選取a[e2]，a[e4]分別作為pivot1，pivot2。由于步驟5進行了排序，所以必有pivot1 <=pivot2。定義兩個指針less和great，less從最左邊開始向右遍歷，一直找到第一個不小于pivot1的元素，great從右邊開始向左遍歷，一直找到第一個不大于pivot2的元素。

		 /*
         * Use the second and fourth of the five sorted elements as pivots.
         * These values are inexpensive approximations of the first and
         * second terciles of the array. Note that pivot1 <= pivot2.
         */
        int pivot1 = a[e2];
        int pivot2 = a[e4];
        /*
         * The first and the last elements to be sorted are moved to the
         * locations formerly occupied by the pivots. When partitioning
         * is complete, the pivots are swapped back into their final
         * positions, and excluded from subsequent sorting.
         */
        a[e2] = a[left];
        a[e4] = a[right];
        /*
         * Skip elements, which are less or greater than pivot values.
         */
        while (a[++less] < pivot1);
        while (a[--great] > pivot2);

7. 接著定義指針k從less-1開始向右遍歷至great，把小于pivot1的元素移動到less左邊，大于pivot2的元素移動到great右邊。這里要注意，我們已知great處的元素小于pivot2，但是它于pivot1的大小關(guān)系，還需要進行判斷，如果比pivot1還小，需要移動到到less左邊，否則只需要交換到k處。

			/*
             * Partitioning:
             *
             *   left part           center part                   right part
             * +--------------------------------------------------------------+
             * |  < pivot1  |  pivot1 <= && <= pivot2  |    ?    |  > pivot2  |
             * +--------------------------------------------------------------+
             *               ^                          ^       ^
             *               |                          |       |
             *              less                        k     great
             *
             * Invariants:
             *
             *              all in (left, less)   < pivot1
             *    pivot1 <= all in [less, k)     <= pivot2
             *              all in (great, right) > pivot2
             *
             * Pointer k is the first index of ?-part.
             */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                short ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /*
                     * Here and below we use "a[i] = b; i++;" instead
                     * of "a[i++] = b;" due to performance issue.
                     */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /*
                     * Here and below we use "a[i] = b; i--;" instead
                     * of "a[i--] = b;" due to performance issue.
                     */
                    a[great] = ak;
                    --great;
                }
            }

8. 將樞軸交換到它們的最終位置

	// Swap pivots into their final positions
    a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
    a[right] = a[great + 1]; a[great + 1] = pivot2;

9. 遞歸排序左右部分，不包括已知的樞軸

		// Sort left and right parts recursively, excluding known pivots
        sort(a, left, less - 2, leftmost);
        sort(a, great + 2, right, false);

10. 對于中間的部分，如果大于4/7的數(shù)組長度，遞歸中間部分

			/*
             * If center part is too large (comprises > 4/7 of the array),
             * swap internal pivot values to ends.
             */
            if (less < e1 && e5 < great) {
                /*
                 * Skip elements, which are equal to pivot values.
                 */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /*
                 * Partitioning:
                 *
                 *   left part         center part                  right part
                 * +----------------------------------------------------------+
                 * | == pivot1 |  pivot1 < && < pivot2  |    ?    | == pivot2 |
                 * +----------------------------------------------------------+
                 *              ^                        ^       ^
                 *              |                        |       |
                 *             less                      k     great
                 *
                 * Invariants:
                 *
                 *              all in (*,  less) == pivot1
                 *     pivot1 < all in [less,  k)  < pivot2
                 *              all in (great, *) == pivot2
                 *
                 * Pointer k is the first index of ?-part.
                 */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    short ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /*
                             * Even though a[great] equals to pivot1, the
                             * assignment a[less] = pivot1 may be incorrect,
                             * if a[great] and pivot1 are floating-point zeros
                             * of different signs. Therefore in float and
                             * double sorting methods we have to use more
                             * accurate assignment a[less] = a[great].
                             */
                            a[less] = pivot1;
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

4 小結(jié)

Arrays.sort對升序數(shù)組、降序數(shù)組和重復(fù)數(shù)組的排序效率有了很大的提升，這里面有幾個重大的優(yōu)化。