归并排序

归并也是经典的分治策略, 它将问题分成一些小的问题后递归求解, 而治就是将分段的各个答案合并补在一起

分解 – 将当前区间一分为二，即求分裂点 mid = (low + high)/2;
求解 – 递归地对两个子区间a[low…mid] 和 a[mid+1…high]进行归并排序。递归的终结条件是子区间长度为1。
合并 – 将已排序的两个子区间a[low…mid]和 a[mid+1…high]归并为一个有序的区间a[low…high]。

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将数组两半进行比较, 较小的放在新的数组中, 最后将没有放完的一半全部放入新的数组中. 最后将新的数组复制给原数组

void merge(int p[], int Frist, int mid, int Last){    int i, j, k;    i = Frist, j = mid + 1, k = 0;    int *a = new int[N];    while (i <= mid && j <= Last)    {        if (p[i] < p[j])            a[k++] = p[i++];        else            a[k++] = p[j++];    }    while (i <= mid)        a[k++] = p[i++];    while (j <= Last)        a[k++] = p[j++];    for (i = 0; i < k; i++)        p[Frist + i] = a[i];    free(a);}

既然要将数组分半, 那就应该用一个 mid = (first + last) / 2;保存中间的位置, 这样将数组分成前一半和后一半, 直到分成一个一个的数

void merge_groups(int a[], int Frist, int Last){    //当数组的 first == last 即分成一个的时候结束分半    if (a == NULL || Frist >= Last)        return;    int mid = (Frist + Last) / 2;    //数组不断地分成两份    merge_groups(a, Frist, mid);    merge_groups(a, mid + 1, Last);    //两个两个数组开始合并    merge(a, Frist, mid, Last);}

源代码

#include <iostream>#include <cstdlib>using namespace std;void merge(int p[], int Frist, int mid, int Last);//void merge_sort(int a[], int n);void merge_groups(int a[], int Frist, int Last);const int N = 20;int main(){    int *p = new int[N];    memset(p, 0, N);    int i = 0, n; cin >> n;    while (i < n)        cin >> p[i++];    //int Frist = 0;    merge_groups(p, 0, n - 1);    for (int i = 0; i < n; i++)        cout << p[i] << " ";    free(p);    system("pause");    return 0;}//void merge_sort(int a[], int n)//{//  if (a == NULL || n <= 0)//      return;//  else//      merge_groups(a, 0, n - 1);//}void merge_groups(int a[], int Frist, int Last){    //当数组的 first == last 即分成一个的时候结束分半    if (a == NULL || Frist >= Last)        return;    int mid = (Frist + Last) / 2;    //数组不断地分成两份    merge_groups(a, Frist, mid);    merge_groups(a, mid + 1, Last);    //两个两个数组开始合并    merge(a, Frist, mid, Last);}void merge(int p[], int Frist, int mid, int Last){    int i, j, k;    i = Frist, j = mid + 1, k = 0;    int *a = new int[N];    while (i <= mid && j <= Last)    {        if (p[i] < p[j])            a[k++] = p[i++];        else            a[k++] = p[j++];    }    while (i <= mid)        a[k++] = p[i++];    while (j <= Last)        a[k++] = p[j++];    for (i = 0; i < k; i++)        p[Frist + i] = a[i];    free(a);}