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.
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
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
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
#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;
}
归并排序的总体复杂度为0(nlogn)
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