「排序算法」专题 4:归并排序的优化
根据上一节的「归并排序」的 3 个优化,写出的代码。
Java 代码:
public class Solution {
/**
* 列表大小等于或小于该大小,将优先于 mergeSort 使用插入排序
*/
private static final int INSERTION_SORT_THRESHOLD = 7;
public int[] sortArray(int[] nums) {
int len = nums.length;
int[] temp = new int[len];
mergeSort(nums, 0, len - 1, temp);
return nums;
}
/**
* 对数组 nums 的子区间 [left, right] 进行归并排序
*
* @param nums
* @param left
* @param right
* @param temp 用于合并两个有序数组的辅助数组,全局使用一份,避免多次创建和销毁
*/
private void mergeSort(int[] nums, int left, int right, int[] temp) {
// 小区间使用插入排序
if (right - left <= INSERTION_SORT_THRESHOLD) {
insertionSort(nums, left, right);
return;
}
int mid = left + (right - left) / 2;
// Java 里有更优的写法,在 left 和 right 都是大整数时,即使溢出,结论依然正确
// int mid = (left + right) >>> 1;
mergeSort(nums, left, mid, temp);
mergeSort(nums, mid + 1, right, temp);
// 如果数组的这个子区间本身有序,无需合并
if (nums[mid] <= nums[mid + 1]) {
return;
}
mergeOfTwoSortedArray(nums, left, mid, right, temp);
}
/**
* 对数组 arr 的子区间 [left, right] 使用插入排序
*
* @param arr 给定数组
* @param left 左边界,能取到
* @param right 右边界,能取到
*/
private void insertionSort(int[] arr, int left, int right) {
for (int i = left + 1; i <= right; i++) {
int temp = arr[i];
int j = i;
while (j > left && arr[j - 1] > temp) {
arr[j] = arr[j - 1];
j--;
}
arr[j] = temp;
}
}
/**
* 合并两个有序数组:先把值复制到临时数组,再合并回去
*
* @param nums
* @param left
* @param mid [left, mid] 有序,[mid + 1, right] 有序
* @param right
* @param temp 全局使用的临时数组
*/
private void mergeOfTwoSortedArray(int[] nums, int left, int mid, int right, int[] temp) {
System.arraycopy(nums, left, temp, left, right + 1 - left);
int i = left;
int j = mid + 1;
for (int k = left; k <= right; k++) {
if (i == mid + 1) {
nums[k] = temp[j];
j++;
} else if (j == right + 1) {
nums[k] = temp[i];
i++;
} else if (temp[i] <= temp[j]) {
// 注意写成 < 就丢失了稳定性(相同元素原来靠前的排序以后依然靠前)
nums[k] = temp[i];
i++;
} else {
// temp[i] > temp[j]
nums[k] = temp[j];
j++;
}
}
}
}
Python 代码:
class MergeSortOptimizer:
def __str__(self):
return "归并排序的优化"
def __merge_of_two_sorted_array(self, arr, left, mid, right):
# 将原数组 [left, right] 区间内的元素复制到辅助数组
for index in range(left, right + 1):
nums_for_compare[index] = arr[index]
i = left
j = mid + 1
for k in range(left, right + 1):
if i == mid + 1:
# i 用完了,就拼命用 j
arr[k] = nums_for_compare[j]
j += 1
elif j > right:
# j 用完了,就拼命用 i
arr[k] = nums_for_compare[i]
i += 1
elif nums_for_compare[i] <= nums_for_compare[j]:
arr[k] = nums_for_compare[i]
i += 1
else:
assert nums_for_compare[i] > nums_for_compare[j]
arr[k] = nums_for_compare[j]
j += 1
def insert_sort_for_sub_interval(self, arr, left, right):
"""多次赋值的插入排序"""
for i in range(left + 1, right + 1):
temp = arr[i]
j = i
# 注意:这里 j 最多到 left
while j > left and arr[j - 1] > temp:
arr[j] = arr[j - 1]
j -= 1
arr[j] = temp
def __merge_sort(self, arr, left, right):
if right - left <= 15:
self.insert_sort_for_sub_interval(arr, left, right)
return
mid = left + (right - left) // 2
self.__merge_sort(arr, left, mid)
self.__merge_sort(arr, mid + 1, right)
if arr[mid] <= arr[mid + 1]:
return
self.__merge_of_two_sorted_array(arr, left, mid, right)
@SortingUtil.cal_time
def sort(self, arr):
global nums_for_compare
size = len(arr)
nums_for_compare = list(range(size))
self.__merge_sort(arr, 0, size - 1)
复杂度分析:
- 时间复杂度:$O(N \log N)$,这里 $N$ 是数组的长度;
- 空间复杂度:$O(N)$,辅助数组与输入数组规模相当。
(本节完)