循環連結清單和約瑟夫環
循環連結清單的實作
單連結清單隻有向後結點,當單連結清單的尾連結清單不指向NULL,而是指向頭結點時候,形成了一個環,成為單循環連結清單,簡稱循環連結清單。當它是空表,向後結點就隻想了自己,這也是它與單連結清單的主要差異,判斷node->next是否等于head。
代碼實作分為四部分:
初始化
插入
删除
定位尋找
代碼實作:
void ListInit(Node *pNode){
int item;
Node *temp,*target;
cout<
while(1){
cin>>item;
if(!item)
return ;
if(!(pNode)){ //當空表的時候,head==NULL
pNode = new Node ;
if(!(pNode))
exit(0);//未成功申請
pNode->data = item;
pNode->next = pNode;
}
else{
//
for(target = pNode;target->next!=pNode;target = target->next)
;
temp = new Node;
if(!(temp))
exit(0);
temp->data = item;
temp->next = pNode;
target->next = temp;
}
}
}
void ListInsert(Node *pNode,int i){ //參數是首節點和插入位置
Node *temp;
Node *target;
int item;
cout<
cin>>item;
if(i==1){
temp = new Node;
if(!temp)
exit(0);
temp->data = item;
for(target=pNode;target->next != pNode;target = target->next)
;
temp->next = pNode;
target->next = temp;
pNode = temp;
}
else{
target = pNode;
for (int j=1;j
target = target->next;
temp = new Node;
if(!temp)
exit(0);
temp->data = item;
temp->next = target->next;
target->next = temp;
}
}
void ListDelete(Node *pNode,int i){
Node *target,*temp;
if(i==1){
for(target=pNode;target->next!=pNode;target=target->next)
;
temp = pNode;//儲存一下要删除的首節點 ,一會便于釋放
pNode = pNode->next;
target->next = pNode;
delete temp;
}
else{
target = pNode;
for(int j=1;j
target = target->next;
temp = target->next;//要釋放的node
target->next = target->next->next;
delete temp;
}
}
int ListSearch(Node *pNode,int elem){ //查詢并傳回結點所在的位置
Node *target;
int i=1;
for(target = pNode;target->data!=elem && target->next!= pNode;++i)
target = target->next;
if(target->next == pNode && target->data!=elem)
return 0;
else return i;
}
約瑟夫問題
約瑟夫環(約瑟夫問題)是一個數學的應用問題:已知n個人(以編号1,2,3...n分别表示)圍坐在一張圓桌周圍。從編号為k的人開始報數,數到m的那個人出列;他的下一個人又從1開始報數,數到m的那個人又出列;依此規律重複下去,直到圓桌周圍的人全部出列。這類問題用循環清單的思想剛好能解決。
注意:編寫代碼的時候,注意報數為m = 1的時候特殊情況
#include
#include
using namespace std;
typedef struct Node{
int data;
Node *next;
};
Node *Create(int n){
Node *p = NULL, *head;
head = new Node;
if (!head)
exit(0);
p = head; // p是目前指針
int item=1;
if(n){
int i=1;
Node *temp;
while(i<=n){
temp = new Node;
if(!temp)
exit(0);
temp->data = i++;
p->next = temp;
p = temp;
}
p->next = head->next;
}
delete head;
return p->next;
}
void Joseph(int n,int m){
//n為總人數,m為數到第m個的退出
m = n%m;
Node *start = Create(n);
if(m){//如果取餘數後的m!=0,說明 m!=1
while(start->next!=start){
Node *temp = new Node;
if(!temp)
exit(0);
for(int i=0;i
start = start->next;
temp = start->next;
start->next = start->next->next;
start = start->next;
cout<data<
delete temp;
}
}
else{
for(int i=0;i
Node *temp = new Node;
if(!temp)
exit(0);
cout<data<
temp = start;
start = start->next;
delete temp;
}
}
cout<
cout<data<
}
int main(){
Joseph(3,1);
Joseph(3,2);
return 0;
}
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