任务描述
本关任务是实现实现二路归并排序算法,void MergeSort(int R[],int N,int low,int high);
void Merge(int R[],int N,int u,int m,int v);
相关知识
为了完成本关任务,你需要掌握:二路归并算法。
归并排序(MERGE-SORT)是建立在归并操作上的一种有效的排序算法,将已有序的子序列合并,得到完全有序的序列。该算法是采用分治法(Divide and Conquer)的一个非常典型的应用。
若将两个有序表合并成一个有序表,称为二路归并。
#include <iostream>
#include <stdio.h>
using namespace std;
#define Max 500 /*N为数据量大小*/
void MergeSort(int R[], int N, int low, int high);
void Merge(int R[], int N, int u, int m, int v);
void Print(int R[], int N);
int main() {
int *R, N, i;
cin >> N;
R = new int[N + 1];
for (i = 1; i <= N; i++)
cin >> R[i];
Print(R, N); // 打印初始数组状态
MergeSort(R, N, 1, N);
Print(R, N); // 打印最终排序后的数组状态
return 0;
}
void Print(int R[], int N) {
int i;
cout << R[1];
for (i = 2; i <= N; i++)
cout << "," << R[i];
cout << endl;
}
void MergeSort(int R[], int N, int low, int high) {
if (low < high) {
int mid = (low + high) / 2;
MergeSort(R, N, low, mid);
MergeSort(R, N, mid + 1, high);
Merge(R, N, low, mid, high);
Print(R, N); // 打印合并后的状态
}
}
void Merge(int R[], int N, int u, int m, int v) {
int n1 = m - u + 1;
int n2 = v - m;
int *L = new int[n1];
int *R2 = new int[n2];
for (int i = 0; i < n1; i++)
L[i] = R[u + i];
for (int j = 0; j < n2; j++)
R2[j] = R[m + 1 + j];
int i = 0, j = 0, k = u;
while (i < n1 && j < n2) {
if (L[i] <= R2[j]) {
R[k++] = L[i++];
} else {
R[k++] = R2[j++];
}
}
while (i < n1) {
R[k++] = L[i++];
}
while (j < n2) {
R[k++] = R2[j++];
}
delete[] L;
delete[] R2;
}