题注:《程序员面试宝典》有相关习题,但思路相对不清晰,排版有错误,本文对此参考相关书籍和自己观点进行了重写,供大家参考。
1.查找链表元素
Step1:置查找标记bfound=false;判断链表是否为空,是,提示“不能查找空链表”;否,进入step2。
Step2:从链表头开始查找,判断(当前点的info是否与待查找元素值相等&&指针未指向末尾),是,“查找结束,bfound=true”,否,继续查找。
Step3:判断bfound= = true,是,“提示查找成功”,否,“提示查找失败”。
/查找单链表元素 template<typename Type> void linkedlistType<Type>::search(const Type& searchItem) { nodeType<Type> *current; bool found = false; if(first == NULL) //1.空链表 { cout << "WARNING: Can not search an empty list!" << endl; return; } else { current = first; while(!found && current != NULL) { if(current->info == searchItem) { found = true; break; } else { current = current->link; } } if(found) { cout << searchItem << " was found in the List! " << endl; } else { cout << searchItem << " was not found in the List! " << endl; } } }
2.删除链表元素值
Step1:置查找标记bfound=false; 判断链表是否为空,是,提示“不能对空链表进行删除操作”;否,进入step2。
Step2:判断待删除元素值是否与头节点元素值相等,是,调整头节点指针;否,进入step3。
Step3:判断链表中是否存在该元素,否,“提示元素不存在”;是,进入step4。
Step4:判定要删除元素是否与末尾节点元素值相等,是,调整末尾last指针;否,此时为中间节点,需要调整trailCurrent和Current指针的指向。
//删除单链表元素 template<typename Type> void linkedlistType<Type>::deleteNode(const Type& deleteItem) { nodeType<Type> *tempNode = new nodeType<Type>; nodeType<Type> *current = new nodeType<Type>; nodeType<Type> *trailCurrent = new nodeType<Type>; bool found; //链表为空 case1 if(first == NULL) { cout << "Can not delete an empty List!" << endl; } else { if( first->info == deleteItem ) { //要删除的也是第一个节点(仅一个节点,或不止一个节点) case2 tempNode = first; first = first->link; if(first == NULL) { last = NULL; } delete tempNode; } else { //先查找,后判断... case3 found = false; trailCurrent = first; current = first->link; while((!found) && (current != NULL)) { if(deleteItem != current->info) { trailCurrent = current; current = current->link; } else { found = true; } } if(found) { //能找到... trailCurrent ->link = current->link; if(current == last) { last = trailCurrent; //case 3a } delete current; //case 3b } //不存在该点...case4 else { cout << "The deleteItem is not Exist in the List! " << endl; } //end else }//end else }//end else }// end deleteNode
3.单链表逆置[迭代实现]
Step1:判断链表是否为空,是,提示“不能对空链表进行逆置操作“;否,进入Step2;
Step2:从第2个节点开始,依次将每个节点插入到第一个节点的前面,判断指针是否指向了链表尾部,是,返回头指针结束;否,继续迭代后面的链表元素。
template<typename Type> nodeType<Type>* linkedlistType<Type>::reverseList() //逆置单链表 { if(first == NULL) { cout << "Can't reverse empty List!" << endl; } else { nodeType<Type>* p = first; nodeType<Type>* q = p->link; while(q != NULL) { p->link = q->link; q->link = first; first = q; q = p->link; } } return first; }
4.单链表排序[直接插入排序]
思路:分为以下几种情况:
1) 单链表为空;
2) 单链表非空,但仅含一个元素,无需排序已经有序;
3) 待插入元素小于头结点的元素;
4) 待插入元素为前已有序的中间的元素值;
5) 待插入的元素前所有元素都比其小,直接插到末尾。
分别用lastInOrder记录已经有序的最后一个节点,firstOutOfOrder第一个尚未排序(正待参与)排序的节点。current用于记录循环的节点,trailCurrent记录current前的节点。
template<typename Type> void linkedlistType<Type>::sortList() //单链表排序 { nodeType<Type>* current; nodeType<Type>* trailCurrent; nodeType<Type>* lastInOrder; nodeType<Type>* firstOutOfOrder; lastInOrder = first; //case1,表为空. if(first == NULL) { cout << "Can't Sort of empty List!" << endl; return; } //case2,表不为空,但表长为1,仅含1个元素. if(first->link == NULL) { cout << "The List Was Already ordered!" << endl; return; } while(lastInOrder->link != NULL) { firstOutOfOrder = lastInOrder->link; //case3,要插入的元素小于第1个元素. if(firstOutOfOrder->info < first->info) { lastInOrder->link = firstOutOfOrder->link; firstOutOfOrder->link = first; first = firstOutOfOrder; } else { trailCurrent = first; current = first->link; while(current->info < firstOutOfOrder->info) { trailCurrent = current; current = current->link; } //case4,要插入的元素在前已有序元素的中间. if(trailCurrent != lastInOrder) { lastInOrder->link = firstOutOfOrder->link; firstOutOfOrder->link = current; trailCurrent->link = firstOutOfOrder; } else { //case5,要插入的元素大于最后一个已经有序的元素. lastInOrder = lastInOrder->link; }//end else }//end else }//end while }
5.单链表在不知道链表长度的前提下求链表中间节点的值。
思路:分以下几种情况:
1) 链表为空;
2) 链表非空,但仅有一个或两个节点;可以直接返回第一个节点的元素值。
3) 链表非空,但含有三个或三个以上的节点,可以通过定义两个指针,一个指针的跳步为2次的时候,另一个指针的跳步为1次,当跳至结尾时,另一个节点恰好在中间位置。
//不知道表长的前提下求单链表中间元素
template<typename Type> Type linkedlistType<Type>::midValOfList() { nodeType<Type> *current; nodeType<Type> *halfCurrent; if(first == NULL) //case1,没有节点 { cout << "链表为空!" << endl; return -1; } else if(first->link == NULL || first->link->link == NULL) //case2,仅一个节点或两个节点. { return first->info; } else //case3,含有三个或三个以上的节点. { current = first; halfCurrent = current; while(current->link != NULL) { current = current->link; if(current->link != NULL) { if(current->link != NULL) { halfCurrent = halfCurrent->link; current = current->link; }//end if } }//end while return halfCurrent->info; }//end else }
6.单链表建立
思路:单链表的建立可分为根据插入新节点的位置的不同而分为两种,1:在链表末尾插入元素的建立方式;2:在链表前面插入元素建立链表的方式。
对应1末尾插入分为两步:
Step1:如果当前链表为空,则置first=last=newNode;否则,进入Step2。
Step2:插入新结点元素,修改last指针。
对于2链表first指针前插入:主要需要保证插入元素后,修正first节点即可。
//正向末尾插入 template<typename Type> nodeType<Type>* linkedlistType<Type>::buildListForward() { nodeType<Type> *newNode; int num; cout << " Enter a list of integer end with -999. " << endl; cin >> num; while(num != -999) { //..add newNode = new nodeType<Type>; newNode->info = num; newNode->link = NULL; if(first==NULL) { first = newNode; last = newNode; } else { last->link = newNode; last = newNode; } cin >> num; } return first; } //反向表头插入,从前面插入... template<typename Type> nodeType<Type>* linkedlistType<Type>::buildListBackward() { nodeType<Type> *newNode; int num; cout << " Enter a list of integer end with -999. " << endl; cin >> num; while(num != -999) { //..add newNode = new nodeType<Type>; newNode->info = num; newNode->link = first; first = newNode; cin >> num; } return first; }
7.单链表的测量长度
思路:链表的长度等效为节点个数,指针非空则循环判断即可。
//求解链表长度 template<typename Type> int linkedlistType<Type>::length() { int count = 0; nodeType<Type> *current; current = first; while(current != NULL) { count++; current = current->link; } return count; //节点个数等效为长度. }
8.单链表的插入
思路:链表的插入也同链表的建立一样分为前向、后向插入两种形式,注意first、last指针的指向问题。
//在前面插入 template<typename Type> void linkedlistType<Type>::insertFirst(const Type& newItem) { //last no use. nodeType<Type> *newNode = new nodeType<Type>; newNode->info = newItem; newNode->link = first; //在前面加入... first = newNode; } //在后面插入元素... template<typename Type> void linkedlistType<Type>::insertLast(const Type& newItem) { nodeType<Type> *newNode = new nodeType<Type>; newNode->info = newItem; newNode->link = NULL; //在后面加入... if(first == NULL) { first = newNode; last = newNode; } else { last->link = newNode; last = newNode; } }
后续陆续会有栈、队列、二叉树、图、排序、查找等的相关分析,希望大家关注!