排序：合并排序

合并排序

1、基本思路

1）分解：将n个元素的数组分成含n/2 个元素的子序列；

2）求解：用合并排序法对每个子序列递归地进行排序；

3）合并：合并已经完成排序子序列。

伪代码如下：

mergesort(A,p,r)    if p<r       q = floor((p+r)/2)       mergesort(A,p,q)       mergesort(A,q+1,r)       merge(A,p,q,r)

merge(A,p,q,r)    n1=q-p+1    n2=r-q    create array L[1,2,...,n1+1] and R[1,2,...,n2+1]    for i=1 to n1         L[i] = A[p+i-1]     for i = 1 to n2         R[i] = A[q+i]    L[n1+1] = +inf    R[n2+1] = +inf    i = 1    j = 1    for k=p to r         if L[i]<=R[j]             A[k] = L[i]             i = i+1         else             A[K] = R[j]             j = j+1 

2、C++代码

#include<limits.h>#include<algorithm>//using namespace std;void merge(int* A, const int p, int q, int r) {//确定两个子数组的长度    int n1 = q - p + 1;int n2 = r - q;//开辟临时内存空间int* L = new int[n1 + 1];int* R = new int[n2 + 1];//将子数组复制到开辟的空间中for (int i = 0; i < n1; ++i)L[i] = A[p + i ];for (int i = 0; i < n2; ++i)R[i] = A[q + i + 1];L[n1 ] = INT_MAX;R[n2 ] = INT_MAX;int i = 0;int j = 0;//合并，类似于两堆牌，比较牌堆顶部的两张牌，小的先放入结果中。for (int k = p; k <= r; ++k) if (L[i] <= R[j])A[k] = L[i++];elseA[k] = R[j++];}void mergesort(int* A, int p, int r) {if (p < r){int q = floor((p + r) / 2);mergesort(A, p, q);mergesort(A, q + 1, r);merge(A, p, q, r);}}int main(){int A[6] = { 5,3,2,4,1,0};int p = 0;int r = 5;mergesort(A, p, r);return 0;}