十大排序算法(七), 归并排序

Posted by zhoujie on April 24, 2021

算法图解

BA0D2CBA-2162-4BA2-A63E-E701D3C8E9C2.png

典型的分治法, 是把排序问题分解到最小单位(即: 1个数排序), 把子树的排序结果向上合成上一层级父亲的排序结果, 下图描述的是的过程

0011886A-933B-4198-949B-2CEC66F10C27.png

945146A9-0DDD-4128-91F7-046736A9811D.png

  1. 先递归对序列进行解成最小单元
  2. 逐级计算的结果

实现

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private static void doSort(int[] arr, int[] tmp, int start, int end) {
        if (start < end) {
            int mid = (end + start) / 2;
            doSort(arr, tmp, start, mid);
            doSort(arr, tmp, mid + 1, end);
            merge(arr, tmp, start, mid, end);
        }
    }

private static void merge(int[] arr, int[] tmp, int start, int mid, int end) {
        int i = start;
        int j = mid + 1;
        int t = 0;
        while (i <= mid && j <= end) {
            if (arr[i] <= arr[j]) {
                tmp[t++] = arr[i++];
            }

            if (arr[i] > arr[j]) {
                tmp[t++] = arr[j++];
            }
        }

        while (i <= mid) {
            tmp[t++] = arr[i++];
        }

        while (j <= end) {
            tmp[t++] = arr[j++];
        }

        System.arraycopy(tmp, 0, arr, start, end - start + 1);
    }

时间复杂度

O(nlogn) 数组被分成了二叉树,二叉树的层高 log2n , 每层需要比对的次数 n f(n) = n * log2n = O(nlogn)

空间复杂度

O(n) 用了一个temp数组,来缓存中间处理的数据, 当数据量很大的时候需要考虑这里的空间浪费问题

完整代码

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class MergeSort {

    private static void sort(int[] array) {
        if (array == null || array.length <= 1) {
            return;
        }

        int[] tmp = new int[array.length];
        int length = array.length;
        doSort(array, tmp, 0, length - 1);
    }

    private static void doSort(int[] arr, int[] tmp, int start, int end) {
        if (start < end) {
            int mid = (end + start) / 2;
            doSort(arr, tmp, start, mid);
            doSort(arr, tmp, mid + 1, end);
            merge(arr, tmp, start, mid, end);
        }
    }

    private static void merge(int[] arr, int[] tmp, int start, int mid, int end){
        int i = start;
        int j = mid + 1;
        int t = 0;
        while (i <= mid && j <= end) {
            if (arr[i] <= arr[j]) {
                tmp[t++] = arr[i++];
            }

            if (arr[i] > arr[j]) {
                tmp[t++] = arr[j++];
            }
        }

        while (i <= mid) {
            tmp[t++] = arr[i++];
        }

        while (j <= end) {
            tmp[t++] = arr[j++];
        }

        System.arraycopy(tmp, 0, arr, start, end - start + 1);
    }

    public static void main(String[] args) {
        int[] array = {111, 52, 77, 98, 36, 12, 12, 48};
        sort(array);
        System.out.println(arrayToString(array));
    }

    private static String arrayToString(int[] array) {
        StringBuilder builder = new StringBuilder();
        for (int t : array) {
            builder.append(t + " ");
        }
        return builder.toString();
    }
}

参考

https://www.cnblogs.com/chengxiao/p/6194356.html



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