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Merge Sort

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废江_小江
发布2022-09-05 13:43:49
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发布2022-09-05 13:43:49
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文章被收录于专栏:总栏目

Question

Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.

代码语言:javascript
复制
Merge(A, left, mid, right)
  n1 = mid - left;
  n2 = right - mid;
  create array L[0...n1], R[0...n2]
  for i = 0 to n1-1
    do L[i] = A[left + i]
  for i = 0 to n2-1
    do R[i] = A[mid + i]
  L[n1] = SENTINEL
  R[n2] = SENTINEL
  i = 0;
  j = 0;
  for k = left to right-1
    if L[i] <= R[j]
      then A[k] = L[i]
           i = i + 1
      else A[k] = R[j]
           j = j + 1
 
Merge-Sort(A, left, right){
  if left+1 < right
    then mid = (left + right)/2;
         call Merge-Sort(A, left, mid)
         call Merge-Sort(A, mid, right)
         call Merge(A, left, mid, right)

Input

In the first line n is given. In the second line, n integers are given.

Output

In the first line, print the sequence S. Two consequtive elements should be separated by a space character.

In the second line, print the number of comparisons.

Constraints

n ≤ 500000

0 ≤ an element in S ≤ 109

Sample Input 1

10

8 5 9 2 6 3 7 1 10 4

Sample Output 1

1 2 3 4 5 6 7 8 9 10

34

Meaning

实现归并排序,除了输出排序后的数组,下一行输出归并排序在比较过程中的次数

Sloution

归并排序利用了分治的思想,首先先分,给数组全部对半分开,分开后分到不能再分,这时候需要合并,写一个合并函数将其合并

Coding

代码语言:javascript
复制
#include<iostream>
using namespace std;
int a[500005];
int tmp[500005];
int  n, cnt;
void Solve(int a[],int l ,int mid ,int r , int tmp[]) {
	//合并,使用三个指针指向三个数组
	int p = 0;
	int p1 = l, p2 = mid + 1;
	while (p1 <= mid && p2 <= r) {
		cnt++;
		if (a[p1] <= a[p2])
			tmp[p++] = a[p1++]; 
		else
			tmp[p++] = a[p2++];
	}
	while (p1 <= mid) {
		cnt++;
		tmp[p++] = a[p1++];
	}
	while (p2 <= r) {
		cnt++;
		tmp[p++] = a[p2++];
	}
	for (int i = 0; i <r-l+1; i++) {
		a[l+i] = tmp[i];
	}
}
void MerageSort(int a[] ,int l ,int r ,int tmp[]) {
	if (l < r) {
		int mid = l + (r - l) / 2;
		MerageSort(a, l, mid, tmp);
		MerageSort(a, mid + 1, r, tmp);
		Solve(a, l, mid, r, tmp);//合并函数
	}
}
int main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	MerageSort(a, 0, n - 1, tmp);
	for (int i = 0; i < n; i++) {
		if (i) cout << " ";
		cout << a[i];
	}
	cout << endl << cnt << endl;
}

Summary

归并排序的总体复杂度为0(nlogn)

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目录
  • Question
  • Meaning
  • Sloution
  • Coding
  • Summary
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