考研系列二-线性表合并

套用系列一的线性表类，又加了三个成员函数：

  1 #include <fstream>
  2 #include <iostream>
  3 #include <cstring>
  4 #include <cstdlib>
  5 
  6 using namespace std;
  7 
  8 #define N 100
  9 enum Status{success,fail,fatal,range_error,overflow};
 10 
 11 class List{
 12     private:
 13     char *list;
 14     int len;
 15 
 16     public:
 17     List(char *tlist=""){
 18         list=new char[N];
 19         strcpy(list,tlist);//初始化
 20         len=strlen(list);
 21     }
 22     ~List(){
 23         delete list;
 24     }
 25     void listClear(){//销毁
 26         list[0]='\0';
 27         len=0;
 28     }
 29     bool listEmpty(){//判空
 30         return len==0;
 31     }
 32     int listSize(){//求长度
 33         return len;
 34     }
 35     Status listRetrieve(int pos,char &elem){//取指定位置元素
 36         if(pos>=0&&pos<len){
 37             elem=list[pos];
 38             return success;
 39         }else   return range_error;
 40     }
 41     Status listLocate(char elem,int &pos);//按值查找
 42     Status listInsert(int pos,char elem);//插入
 43     Status listRemove(int pos);//删除
 44     Status listPrior(int pos,char &elem);//求前驱
 45     Status listNext(int pos,char &elem);//求后继
 46     void listShow();//显示表内容
 47     Status listUnion(List &list2);//集合合并
 48     Status listMerge(List &list1,List &list2);//有序表合并
 49 };
 50 
 51 int main(){
 52     //freopen("D:\\input.in","r",stdin);
 53     //freopen("D:\\output.out","w",stdout);
 54     List s1("hello"),s2("echo"),s10("abcdegkl"),s11("bdefhk"),s12;
 55 
 56     s1.listShow();
 57     if(s1.listUnion(s2)==success)   s1.listShow();
 58 
 59     s10.listShow();
 60     s11.listShow();
 61     if(s12.listMerge(s10,s11)==success)     s12.listShow();
 62     return 0;
 63 }
 64 Status List::listLocate(char elem,int &pos){
 65     for(int i=0;i<len;i++){
 66         if(list[i]==elem){
 67             pos=i;
 68             return success;
 69         }
 70     }
 71     return fail;
 72 }
 73 Status List::listInsert(int pos,char elem){
 74     if(len==N-1)     return overflow;//因为终止符
 75     if(pos<0||pos>len)  return range_error;
 76     for(int i=len+1;i>pos;i--)    list[i]=list[i-1];
 77     list[pos]=elem;
 78     len++;
 79     return success;
 80 }
 81 Status List::listRemove(int pos){
 82     if(pos<0||pos>=len)  return range_error;
 83     for(int i=pos;i<len-1;i++)    list[i]=list[i+1];
 84     len--;
 85     return success;
 86 }
 87 Status List::listPrior(int pos,char &elem){
 88     return listRetrieve(pos-1,elem);
 89 }
 90 Status List::listNext(int pos,char &elem){
 91     return listRetrieve(pos+1,elem);
 92 }
 93 void List::listShow(){
 94     puts(list);
 95 }
 96 Status List::listUnion(List &list2){
 97     char elem;
 98     Status status;
 99     int pos,len2=list2.listSize();
100     for(int i=0;i<len2;i++){
101         list2.listRetrieve(i,elem);
102         status=listLocate(elem,pos);
103         if(status!=success){
104             status=listInsert(len,elem);
105             if(status!=success) break;
106         }
107     }
108     return status;
109 }
110 Status List::listMerge(List &list1,List &list2){
111     char elem1,elem2;
112     int len1,len2,i=0,j=0;
113     Status status;
114     len1=list1.listSize();
115     len2=list2.listSize();
116     while(i<len1&&j<len2){
117         list1.listRetrieve(i,elem1);
118         list2.listRetrieve(j,elem2);
119         if(elem1<elem2){
120             status=listInsert(len,elem1);
121             i++;
122         }else{
123             status=listInsert(len,elem2);
124             j++;
125         }
126         if(status!=success) return status;
127     }
128     while(i<len1){
129         list1.listRetrieve(i,elem1);
130         listInsert(len,elem1);
131         if(status!=success) return status;
132         i++;
133     }
134     while(j<len2){
135         list2.listRetrieve(j,elem2);
136         status=listInsert(len,elem2);
137         if(status!=success) return status;
138         j++;
139     }
140     return status;
141 }

敲代码期间遇到一个bug，就是字符串的终止符问题，合并时不小心把终止符给覆盖了，结果程序就飞了。

转载于:https://www.cnblogs.com/jiu0821/p/4923021.html