常见的链表题目总结

考研结束了~虽然不甘心~但是确实输了~想起那段狂啃书本的日子~确实………………

好了~总结一下复习数据结构这门课时做到的链表题目，可能对各位patner都会有用吧~也借此向我那段时光做个了结吧……

//首先对于单链表的数据结构 typedef struct Node { elemType data; //存储节点信息 Type otherInfo; //其它信息 struct Node* next; }Node,*LinkList; 

 

1.单链表的逆序

void Reverse(LinkList& head) { if(NULL == head || NULL == head->next) { return head; } Node* p = head->next; Node* pre = NULL; head->next = NULL; while(p != NULL) { pre = p->next; p->next = head->next; p = pre; } } 

 

2.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(LinkList& head, int num) { Node *cur = head; Node *tail = NULL; int i = 0; if(num <= 0 || NULL ==head) { return FALSE; } for(i = 1; i < num; ++i) { if(NULL == cur) { return FALS; } 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

//假设为Y型相交 //自己可以在纸上画一下，可以很容易的看出，最后一个节点是两个链表共有的 BOOL detect_looplink(LinkList LA,LinkList LB) { if(NULL == LA || NULL == LB || NULL == LA->next || NULL == LB->next) { return FALSE; } Node* pA = LA; Node* pB = LB; while(pA->next) { pA = pA->next; } while(pB->next) { pB = pB->next; } if(pA == pB) { return TRUE; } return FALSE; }  

 

4.检测两个相交链表的交点

//假设为Y型相交 #include <math.h> Node* get_looplinknode(LinkList L,LinkList) { if(!detect_looplink(LA,LB)) { return NULL; } int iLengthA = 0; int iLengthB = 0; Node* p = LA->next; Node* q = LB->next; while(p) { iLengthA++; p = p->next; } while(q) { iLengthB++; q = q->next; } p = LA->next; if(iLengthA > iLengthB) { p = LA->next; q = LB->next; int offset = iLengthA - iLengthB; int i = 0; while(i < offset) { p = p->next; } while(p != q) { p = p->next; q = q->next; } } else { p = LA->next; q = LB->next; int offset = iLengthB - iLengthB; int i = 0; while(i < offset) { q = q->next; } while(p != q) { p = p->next; q = q->next; } } return p; } 

 

5.从一个无头链表中删除节点（该节点不是头、尾节点）

void delete_node(Node* list) { Node* p = list->next; Swap(list->data,p->data); list->next = p->next; free(p); } 

 

6.单链表的逆序输出（递归）

void reverse_print(LinkList head) { if(head) { reverse_print(head->next); print(head->data) } } 

 

7.判断一个单链表是否有环

BOOL islooplink(LinkList head) { Node* slow = head; Node* fast = head; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; if(slow == fast) break; } retun !(fast == NULL || fast->next == NULL); } 

 

8.找出一个有环的单链表的环入口点

/*让我们先来看分析： 当fast若与slow相遇时，slow肯定没有走遍历完链表，而fast已经在环内循环了n圈(1<=n)。假设slow走了s步，则 fast走了2s步（fast步数还等于s 加上在环上多转的n圈），设环长为r，则： 2s = s + nr s= nr 设整个链表长L，入口环与相遇点距离为x，起点到环入口点的距离为a。 a + x = nr a + x = (n – 1)r +r = (n-1)r + L - a a = (n-1)r + (L – a – x) (L – a – x)为相遇点到环入口点的距离，由此可知，从链表头到环入口点等于(n-1)循环内环+相遇点到环入口点，于是我们从链表头、与相遇点分别设一个指针，每 次各走一步，两个指针必定相遇，且相遇第一点为环入口点*/ Node* find_loopnod(LinkList head) { Node* slow = head; Node* fast = head; while ( fast && fast->next ) { slow = slow->next; fast = fast->next->next; if ( slow == fast ) break; } if (fast == NULL || fast->next == NULL) return NULL; slow = head; while (slow != fast) { slow = slow->next; fast = fast->next; } return slow; } 

好了~暂时这么多~以后如果有还会不断更新的~呵呵~