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.
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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
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#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）
