数据结构之两个线性表的合并

           严蔚敏版的教材，算法2.2，下面是代码实现。。。。

#include <iostream>using namespace std;typedef int element;#define maxsize 100typedef struct{  element aa[maxsize];  int last;}sqlist;void mergelist(sqlist *la,sqlist *lb,sqlist *lc){  int i=0,j=0,k=0;  while((i<=la->last)&&(j<=lb->last))  {  if(la->aa[i]<=lb->aa[j])  {    lc->aa[k]=la->aa[i];i++;k++;  }  else  {    lc->aa[k]=lb->aa[j];j++;k++;  }  }  while(i<=la->last)  {  lc->aa[k]=la->aa[i];  i++;  k++;  }  while(j<=lb->last)  {    lc->aa[k]=lb->aa[j];j++;k++;  }  lc->last=la->last+lb->last+1;}int main(){int len,i;  sqlist *la,*lb,*lc;  la=(sqlist*)malloc(sizeof(sqlist));    cout<<"输入链表a的长度:"<<endl;  cin>>len;  la->last=len-1;  cout<<"输入元素"<<endl;  for(i=0;i<len;++i)  cin>>la->aa[i]; lb=(sqlist*)malloc(sizeof(sqlist));    cout<<"输入链表b的长度:"<<endl;  cin>>len;  lb->last=len-1;  cout<<"输入元素"<<endl;  for(i=0;i<len;++i)  cin>>lb->aa[i];  lc=(sqlist*)malloc(sizeof(sqlist));  mergelist(la,lb,lc);  for(i=0;i<=lc->last;++i)  cout<<lc->aa[i]<<"  ";  cout<<endl;  return 0；

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