常见的链表题目

一些常见的单链表题目,总结思路和实现代码。

1.单链表的反序

2.给单链表建环

3.检测单链表是否有环

4.给单链表解环

5.检测两条链表是否相交

6.不输入头节点,删除单链表的指定节点(只给定待删除节点指针)

 

1.单链表的反序

//逆转链表,并返回逆转后的头节点 node* reverse(node *head) { if(head == NULL || head->next == NULL) { return head; } node *cur = head; node *pre = NULL; node *tmp; while(cur->next) { tmp = pre; pre = cur; cur = cur->next; pre->next = tmp; //操作pre的next逆转 } cur->next = pre; //结束时,操作cur的next逆转 return cur; }

 

2.给单链表建环

 

//给单链表建环,让尾指针,指向第num个节点,若没有,返回false bool bulid_looplink(node *head, int num) { node *cur = head; node *tail = NULL; int i = 0; if(num <= 0 || head == NULL) { return false; } for(i = 1; i < num; ++i) { if(cur == NULL) { return false; } cur = cur->next; } tail = cur; while(tail->next) { tail = tail->next; } tail->next = cur; return true; }

 

3.检测单链表是否有环

//检测单链表是否有环,快慢指针 bool detect_looplink(node *head) { node *quick_node = head->next, *slow_node = head; if(head == NULL || head->next == NULL) { return false; } while(quick_node != slow_node) { if(quick_node == NULL || slow_node == NULL) break; quick_node = quick_node->next->next; slow_node = slow_node->next; } if(quick_node != NULL && slow_node != NULL) //非尾节点相遇 return true; return false; }

 

4.给单链表解环

ps:为了增加节点位图的效率,本应使用hash或则红黑树,这里不造车了,直接用 set容器

//找到有环节点,并解环,找到并解环,返回true,无环,返回false //思路:先找到环节点:被2个节点指向的节点(一定有环的条件)ps:不考虑中间环,因为只有一个next节点,只可能是尾环 bool unloop_link(node *head) { set node_bitmap; //node的地址位图 unsigned int num = 0; node *cur = head, *pre = NULL; while(cur != NULL) { if(!node_bitmap.count(cur) ) //该节点未被遍历过 { node_bitmap.insert(cur); ++num; } else //指向已被遍历过的节点,此时pre节点为尾节点 { pre->next = NULL; return true; } pre = cur; cur = cur->next; } return false; }

 

5.检测两条链表是否相交

//检测两条链表是否相交,是则返回第一个交点,否则返回NULL //思路:把2个链表各遍历一遍,记下长度length1和length2,若2者的尾节点指针相等,则相交。 // 之后再把长的链表从abs(len1-len2)的位置开始遍历,第一个相等的指针为目标节点 node* detect_intersect_links(node *first_link, node *second_link) { int legnth1 = 1, length2 = 1, pos = 0; node *cur = NULL, *longer_link = first_link, *shorter_link = second_link; if(first_link == NULL || second_link == NULL) { return NULL; } while(first_link->next || second_link->next) //遍历2个链表 { if(first_link->next) { first_link = first_link->next; ++legnth1; } if(second_link->next) { second_link = second_link->next; ++length2; } } if(first_link != second_link) //比较尾节点 { return NULL; } pos = legnth1 - length2; if(legnth1 < length2) //保证 longer_link为长链表 { pos = length2 - legnth1; cur = longer_link; longer_link = shorter_link; shorter_link = cur; } while(pos-- > 0) longer_link = longer_link->next; while(longer_link || shorter_link) { if(longer_link == shorter_link) //找到第一个交点 { return longer_link; } longer_link = longer_link->next; shorter_link = shorter_link->next; } return NULL; }

6.不输入头节点,删除单链表的指定节点(只给定待删除节点指针)

//无头节点,随机给出单链表中一个非头节点,删除该节点,当传入空节点,或者尾节点时,返回false //思路:由于没有头节点,非循环单链表,无法获取目标节点的前节点,所以只能把它的next节点数据前移,并删除next节点 //ps:当传入节点为尾节点,无法用此方法删除 bool withouthead_delete_node(node *target_node) { node *cur = NULL; if(target_node == NULL || target_node->next == NULL) //空节点或者尾节点,失败 { return false; } cur = target_node->next; target_node->name = cur->name; target_node->next = cur->next; delete cur; return true; }

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