11.6 Merge Sort¶
Merge sort is a sorting algorithm based on a divide-and-conquer strategy, consisting of the "divide" and "merge" phases shown in Figure 11-10.
- Divide phase: Recursively split the array at the midpoint, reducing the problem of sorting a long array to the problem of sorting shorter arrays.
- Merge phase: When a sub-array has length 1, stop dividing and start merging, continuously combining the shorter sorted sub-arrays on the left and right into a longer sorted array until the process is complete.
Figure 11-10 Divide and merge phases of merge sort
11.6.1 Algorithm Flow¶
As shown in Figure 11-11, the "divide phase" recursively splits the array from the midpoint into two sub-arrays from top to bottom.
- Calculate the array midpoint
mid, recursively divide the left sub-array (interval[left, mid]) and right sub-array (interval[mid + 1, right]). - Repeat step
1.recursively until a sub-array has length 1.
The "merge phase" merges the left and right sub-arrays into a sorted array from bottom to top. Note that merging starts from sub-arrays of length 1, so every sub-array involved in this phase is already sorted.
Figure 11-11 Merge sort steps
The recursive order of merge sort is consistent with the post-order traversal of a binary tree.
- Post-order traversal: First recursively traverse the left subtree, then recursively traverse the right subtree, and finally process the root node.
- Merge sort: First recursively process the left sub-array, then recursively process the right sub-array, and finally perform the merge.
The implementation of merge sort is shown in the code below. Note that the interval to be merged in nums is [left, right], while the corresponding interval in tmp is [0, right - left].
merge_sort.py
def merge(nums: list[int], left: int, mid: int, right: int):
"""Merge left subarray and right subarray"""
# Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
# Create a temporary array tmp to store the merged results
tmp = [0] * (right - left + 1)
# Initialize the start indices of the left and right subarrays
i, j, k = left, mid + 1, 0
# While both subarrays still have elements, compare and copy the smaller element into the temporary array
while i <= mid and j <= right:
if nums[i] <= nums[j]:
tmp[k] = nums[i]
i += 1
else:
tmp[k] = nums[j]
j += 1
k += 1
# Copy the remaining elements of the left and right subarrays into the temporary array
while i <= mid:
tmp[k] = nums[i]
i += 1
k += 1
while j <= right:
tmp[k] = nums[j]
j += 1
k += 1
# Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for k in range(0, len(tmp)):
nums[left + k] = tmp[k]
def merge_sort(nums: list[int], left: int, right: int):
"""Merge sort"""
# Termination condition
if left >= right:
return # Terminate recursion when subarray length is 1
# Divide and conquer stage
mid = (left + right) // 2 # Calculate midpoint
merge_sort(nums, left, mid) # Recursively process the left subarray
merge_sort(nums, mid + 1, right) # Recursively process the right subarray
# Merge stage
merge(nums, left, mid, right)
merge_sort.cpp
/* Merge left subarray and right subarray */
void merge(vector<int> &nums, int left, int mid, int right) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
vector<int> tmp(right - left + 1);
// Initialize the start indices of the left and right subarrays
int i = left, j = mid + 1, k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j])
tmp[k++] = nums[i++];
else
tmp[k++] = nums[j++];
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmp.size(); k++) {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
void mergeSort(vector<int> &nums, int left, int right) {
// Termination condition
if (left >= right)
return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
int mid = left + (right - left) / 2; // Calculate midpoint
mergeSort(nums, left, mid); // Recursively process the left subarray
mergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.java
/* Merge left subarray and right subarray */
void merge(int[] nums, int left, int mid, int right) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
int[] tmp = new int[right - left + 1];
// Initialize the start indices of the left and right subarrays
int i = left, j = mid + 1, k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j])
tmp[k++] = nums[i++];
else
tmp[k++] = nums[j++];
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmp.length; k++) {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
void mergeSort(int[] nums, int left, int right) {
// Termination condition
if (left >= right)
return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
int mid = left + (right - left) / 2; // Calculate midpoint
mergeSort(nums, left, mid); // Recursively process the left subarray
mergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.cs
/* Merge left subarray and right subarray */
void Merge(int[] nums, int left, int mid, int right) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
int[] tmp = new int[right - left + 1];
// Initialize the start indices of the left and right subarrays
int i = left, j = mid + 1, k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j])
tmp[k++] = nums[i++];
else
tmp[k++] = nums[j++];
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmp.Length; ++k) {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
void MergeSort(int[] nums, int left, int right) {
// Termination condition
if (left >= right) return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
int mid = left + (right - left) / 2; // Calculate midpoint
MergeSort(nums, left, mid); // Recursively process the left subarray
MergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
Merge(nums, left, mid, right);
}
merge_sort.go
/* Merge left subarray and right subarray */
func merge(nums []int, left, mid, right int) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
tmp := make([]int, right-left+1)
// Initialize the start indices of the left and right subarrays
i, j, k := left, mid+1, 0
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
for i <= mid && j <= right {
if nums[i] <= nums[j] {
tmp[k] = nums[i]
i++
} else {
tmp[k] = nums[j]
j++
}
k++
}
// Copy the remaining elements of the left and right subarrays into the temporary array
for i <= mid {
tmp[k] = nums[i]
i++
k++
}
for j <= right {
tmp[k] = nums[j]
j++
k++
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for k := 0; k < len(tmp); k++ {
nums[left+k] = tmp[k]
}
}
/* Merge sort */
func mergeSort(nums []int, left, right int) {
// Termination condition
if left >= right {
return
}
// Divide and conquer stage
mid := left + (right - left) / 2
mergeSort(nums, left, mid)
mergeSort(nums, mid+1, right)
// Merge stage
merge(nums, left, mid, right)
}
merge_sort.swift
/* Merge left subarray and right subarray */
func merge(nums: inout [Int], left: Int, mid: Int, right: Int) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
var tmp = Array(repeating: 0, count: right - left + 1)
// Initialize the start indices of the left and right subarrays
var i = left, j = mid + 1, k = 0
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while i <= mid, j <= right {
if nums[i] <= nums[j] {
tmp[k] = nums[i]
i += 1
} else {
tmp[k] = nums[j]
j += 1
}
k += 1
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while i <= mid {
tmp[k] = nums[i]
i += 1
k += 1
}
while j <= right {
tmp[k] = nums[j]
j += 1
k += 1
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for k in tmp.indices {
nums[left + k] = tmp[k]
}
}
/* Merge sort */
func mergeSort(nums: inout [Int], left: Int, right: Int) {
// Termination condition
if left >= right { // Terminate recursion when subarray length is 1
return
}
// Divide and conquer stage
let mid = left + (right - left) / 2 // Calculate midpoint
mergeSort(nums: &nums, left: left, right: mid) // Recursively process the left subarray
mergeSort(nums: &nums, left: mid + 1, right: right) // Recursively process the right subarray
// Merge stage
merge(nums: &nums, left: left, mid: mid, right: right)
}
merge_sort.js
/* Merge left subarray and right subarray */
function merge(nums, left, mid, right) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
const tmp = new Array(right - left + 1);
// Initialize the start indices of the left and right subarrays
let i = left,
j = mid + 1,
k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j]) {
tmp[k++] = nums[i++];
} else {
tmp[k++] = nums[j++];
}
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmp.length; k++) {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
function mergeSort(nums, left, right) {
// Termination condition
if (left >= right) return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
let mid = Math.floor(left + (right - left) / 2); // Calculate midpoint
mergeSort(nums, left, mid); // Recursively process the left subarray
mergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.ts
/* Merge left subarray and right subarray */
function merge(nums: number[], left: number, mid: number, right: number): void {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
const tmp = new Array(right - left + 1);
// Initialize the start indices of the left and right subarrays
let i = left,
j = mid + 1,
k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j]) {
tmp[k++] = nums[i++];
} else {
tmp[k++] = nums[j++];
}
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmp.length; k++) {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
function mergeSort(nums: number[], left: number, right: number): void {
// Termination condition
if (left >= right) return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
let mid = Math.floor(left + (right - left) / 2); // Calculate midpoint
mergeSort(nums, left, mid); // Recursively process the left subarray
mergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.dart
/* Merge left subarray and right subarray */
void merge(List<int> nums, int left, int mid, int right) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
List<int> tmp = List.filled(right - left + 1, 0);
// Initialize the start indices of the left and right subarrays
int i = left, j = mid + 1, k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j])
tmp[k++] = nums[i++];
else
tmp[k++] = nums[j++];
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmp.length; k++) {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
void mergeSort(List<int> nums, int left, int right) {
// Termination condition
if (left >= right) return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
int mid = left + (right - left) ~/ 2; // Calculate midpoint
mergeSort(nums, left, mid); // Recursively process the left subarray
mergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.rs
/* Merge left subarray and right subarray */
fn merge(nums: &mut [i32], left: usize, mid: usize, right: usize) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
let tmp_size = right - left + 1;
let mut tmp = vec![0; tmp_size];
// Initialize the start indices of the left and right subarrays
let (mut i, mut j, mut k) = (left, mid + 1, 0);
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while i <= mid && j <= right {
if nums[i] <= nums[j] {
tmp[k] = nums[i];
i += 1;
} else {
tmp[k] = nums[j];
j += 1;
}
k += 1;
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while i <= mid {
tmp[k] = nums[i];
k += 1;
i += 1;
}
while j <= right {
tmp[k] = nums[j];
k += 1;
j += 1;
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for k in 0..tmp_size {
nums[left + k] = tmp[k];
}
}
/* Merge sort */
fn merge_sort(nums: &mut [i32], left: usize, right: usize) {
// Termination condition
if left >= right {
return; // Terminate recursion when subarray length is 1
}
// Divide and conquer stage
let mid = left + (right - left) / 2; // Calculate midpoint
merge_sort(nums, left, mid); // Recursively process the left subarray
merge_sort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.c
/* Merge left subarray and right subarray */
void merge(int *nums, int left, int mid, int right) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
int tmpSize = right - left + 1;
int *tmp = (int *)malloc(tmpSize * sizeof(int));
// Initialize the start indices of the left and right subarrays
int i = left, j = mid + 1, k = 0;
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j]) {
tmp[k++] = nums[i++];
} else {
tmp[k++] = nums[j++];
}
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (k = 0; k < tmpSize; ++k) {
nums[left + k] = tmp[k];
}
// Free memory
free(tmp);
}
/* Merge sort */
void mergeSort(int *nums, int left, int right) {
// Termination condition
if (left >= right)
return; // Terminate recursion when subarray length is 1
// Divide and conquer stage
int mid = left + (right - left) / 2; // Calculate midpoint
mergeSort(nums, left, mid); // Recursively process the left subarray
mergeSort(nums, mid + 1, right); // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right);
}
merge_sort.kt
/* Merge left subarray and right subarray */
fun merge(nums: IntArray, left: Int, mid: Int, right: Int) {
// Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
// Create a temporary array tmp to store the merged results
val tmp = IntArray(right - left + 1)
// Initialize the start indices of the left and right subarrays
var i = left
var j = mid + 1
var k = 0
// While both subarrays still have elements, compare and copy the smaller element into the temporary array
while (i <= mid && j <= right) {
if (nums[i] <= nums[j])
tmp[k++] = nums[i++]
else
tmp[k++] = nums[j++]
}
// Copy the remaining elements of the left and right subarrays into the temporary array
while (i <= mid) {
tmp[k++] = nums[i++]
}
while (j <= right) {
tmp[k++] = nums[j++]
}
// Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
for (l in tmp.indices) {
nums[left + l] = tmp[l]
}
}
/* Merge sort */
fun mergeSort(nums: IntArray, left: Int, right: Int) {
// Termination condition
if (left >= right) return // Terminate recursion when subarray length is 1
// Divide and conquer stage
val mid = left + (right - left) / 2 // Calculate midpoint
mergeSort(nums, left, mid) // Recursively process the left subarray
mergeSort(nums, mid + 1, right) // Recursively process the right subarray
// Merge stage
merge(nums, left, mid, right)
}
merge_sort.rb
### Merge left and right subarrays ###
def merge(nums, left, mid, right)
# Left subarray interval is [left, mid], right subarray interval is [mid+1, right]
# Create temporary array tmp to store merged result
tmp = Array.new(right - left + 1, 0)
# Initialize the start indices of the left and right subarrays
i, j, k = left, mid + 1, 0
# While both subarrays still have elements, compare and copy the smaller element into the temporary array
while i <= mid && j <= right
if nums[i] <= nums[j]
tmp[k] = nums[i]
i += 1
else
tmp[k] = nums[j]
j += 1
end
k += 1
end
# Copy the remaining elements of the left and right subarrays into the temporary array
while i <= mid
tmp[k] = nums[i]
i += 1
k += 1
end
while j <= right
tmp[k] = nums[j]
j += 1
k += 1
end
# Copy the elements from the temporary array tmp back to the original array nums at the corresponding interval
(0...tmp.length).each do |k|
nums[left + k] = tmp[k]
end
end
### Merge sort ###
def merge_sort(nums, left, right)
# Termination condition
# Terminate recursion when subarray length is 1
return if left >= right
# Divide and conquer stage
mid = left + (right - left) / 2 # Calculate midpoint
merge_sort(nums, left, mid) # Recursively process the left subarray
merge_sort(nums, mid + 1, right) # Recursively process the right subarray
# Merge stage
merge(nums, left, mid, right)
end
11.6.2 Algorithm Characteristics¶
- Time complexity is \(O(n \log n)\); merge sort is non-adaptive: The divide phase produces a recursion tree of height \(\log n\), and the total number of operations performed during merging at each level is \(n\), so the overall time complexity is \(O(n \log n)\).
- Space complexity is \(O(n)\); merge sort is not in-place: The recursion depth is \(\log n\), which uses \(O(\log n)\) stack-frame space. The merge operation requires an auxiliary array, which uses \(O(n)\) additional space.
- Stable sort: During merging, the relative order of equal elements remains unchanged.
11.6.3 Linked List Sorting¶
For linked lists, merge sort has significant advantages over other sorting algorithms, and it can reduce the space complexity of the sorting task to \(O(1)\).
- Divide phase: Iteration can be used instead of recursion to split the linked list, thereby eliminating the stack-frame space used by recursion.
- Merge phase: In linked lists, node insertion and deletion require only pointer updates, so the merge phase (merging two short sorted linked lists into one longer sorted linked list) does not require creating an additional linked list.
The specific implementation details are quite complex, and interested readers can consult related materials for learning.