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;
}