基础算法之归并排序

传统归并排序如下:

    public static final int[] arr = {};

    private static void sortProcess(int[] arr, int L, int R) {
        if (L == R) {
            return;
        }
        //L和R中点的位置
        int mid = L + ((R - L) >> 1);
        //T(N/2)
        sortProcess(arr, L, mid);
        //T(N/2)
        sortProcess(arr, mid + 1, R);
        //T(N) = 2(T(N/2)) + O(N)  =  O(N*logN)
        merge(arr, L, mid, R);
    }

   private static void merge(int[] arr, int L, int mid, int R) {
        int[] help = new int[R - L + 1];
        int i = 0;
        int p1 = L;
        int p2 = mid + 1;
        while (p1 <= mid && p2 <= R) {
            // 从小到大赋值到help数组上
            help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
        }
        //两个必有且只有一个越界
        while (p1 <= mid) {
            help[i++] = arr[p1++];
        }
        while (p2 <= R) {
            help[i++] = arr[p2++];
        }
        for (i = 0; i < help.length; i++) {
            arr[L + i] = help[i];
        }
    }

时间复杂度为O(n * logn)

非递归版本

public static final int[] arr = {};
    public static void mergeSort2(int n) {
        // 一共发生O(logn)次
        for(int l, m, r, step = 1; step < n; step <<=1 ) {
            // 内部分组merge, 时间复杂度O(n)
            l = 0;
            while (l < n) {
                m = l + step - 1;
                if (m + 1 >= n) {
                    // 已经没有右侧了
                    break;
                }
                // 有右侧，求右侧的右边界
                r = Math.min(l + (step << 1) - 1, n - 1);
                merge(l, m, r);
                l = r + 1;
            }
        }
    }

  public static void merge(int L, int mid, int R) {
        int[] help = new int[R - L + 1];
        int i = 0;
        int p1 = L;
        int p2 = mid + 1;
        while (p1 <= mid && p2 <= R) {
            help[i++] =
                    arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
        }
        //两个必有且只有一个越界
        while (p1 <= mid) {
            help[i++] = arr[p1++];
        }
        while (p2 <= R) {
            help[i++] = arr[p2++];
        }
        for (i = 0; i < help.length; i++) {
            arr[L + i] = help[i];
        }
    }