Merge sort is a sorting algorithm based on the divide-and-conquer strategy, involving the “divide” and “merge” phases shown in the figure below.
- Divide phase: Recursively split the array from the midpoint, transforming the sorting problem of a long array into that of shorter arrays.
- Merge phase: Stop dividing when the length of the sub-array is 1, start merging, and continuously combine two shorter ordered arrays into one longer ordered array until the process is complete.
Algorithm workflow
As shown in the figure below, the “divide phase” recursively splits the array from the midpoint into two sub-arrays from top to bottom.
- Calculate the midpoint
mid
, recursively divide the left sub-array (interval[left, mid]
) and the right sub-array (interval[mid + 1, right]
). - Continue with step
1.
recursively until the sub-array interval length is 1 to stop.
The “merge phase” combines the left and right sub-arrays into a single ordered array from bottom to top. Note that merging starts with sub-arrays of length 1, and each sub-array is ordered during the merge phase.
=== “<1>”
=== “<2>”
=== “<3>”
=== “<4>”
=== “<5>”
=== “<6>”
=== “<7>”
=== “<8>”
=== “<9>”
=== “<10>”
It is observed that the order of recursion in merge sort is consistent with the post-order traversal of a binary tree.
- Post-order traversal: First recursively traverse the left subtree, then the right subtree, and finally handle the root node.
- Merge sort: First recursively handle the left sub-array, then the right sub-array, and finally perform the merge.
The implementation of merge sort is shown in the following code. Note that the interval to be merged in nums
is [left, right]
, while the corresponding interval in tmp
is [0, right - left]
.
Java Code
public static void merge(int[] nums, int low, int mid, int high) {
int left = low;
int right = mid + 1;
List<Integer> ans = new ArrayList<>();
while (left <= mid && right <= high) {
if (nums[left] <= nums[right]) {
ans.add(nums[left]);
left++;
} else {
ans.add(nums[right]);
right++;
}
}
while (left <= mid) {
ans.add(nums[left]);
left++;
}
while (right <= high) {
ans.add(nums[right]);
right++;
}
// move all elements to the original array
for (int i = low; i <= high; i++) {
nums[i] = ans.get(i - low);
}
}
private static void mergeSort(int[] nums, int low, int high) {
if (low >= high) {
return;
}
int mid = (low + high) / 2;
mergeSort(nums, low, mid);
mergeSort(nums, mid + 1, high);
merge(nums, low, mid, high);
}
Python Code
class Solution:
def merge(arr, low, mid, high):
left = low
right = mid + 1
temp = []
while left <= mid and right <= high:
if arr[left] <= arr[right]:
temp.append(arr[left])
left += 1
else:
temp.append(arr[right])
right += 1
while left <= mid:
temp.append(arr[left])
left += 1
while right <= high:
temp.append(arr[right])
right += 1
for i in range(low, high + 1):
arr[i] = temp[i - low]
def mergeSort(arr, low, high):
if low >= high:
return
mid = (low + high) // 2
Solution.mergeSort(arr, low, mid)
Solution.mergeSort(arr, mid + 1, high)
Solution.merge(arr, low, mid, high)
def sortArray(self, nums: List[int]) -> List[int]:
Solution.mergeSort(nums, 0, len(nums) - 1)
return nums
Algorithm characteristics
- Time complexity of $O(n \log n)$, non-adaptive sort: The division creates a recursion tree of height $\log n$, with each layer merging a total of $n$ operations, resulting in an overall time complexity of $O(n \log n)$.
- Space complexity of $O(n)$, non-in-place sort: The recursion depth is $\log n$, using $O(\log n)$ stack frame space. The merging operation requires auxiliary arrays, using an additional space of $O(n)$.
- Stable sort: During the merging process, the order of equal elements remains unchanged.
Ref :-