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1.判断单链表是否有环
使用两个slow, fast指针从头开始扫描链表。指针slow 每次走1步,指针fast每次走2步。如果存在环,则指针slow、fast会相遇;如果不存在环,指针fast遇到NULL退出。
就是所谓的追击相遇问题:

2.求有环单链表的环长
在环上相遇后,记录第一次相遇点为Pos,之后指针slow继续每次走1步,fast每次走2步。在下次相遇的时候fast比slow正好又多走了一圈,也就是多走的距离等于环长。
设从第一次相遇到第二次相遇,设slow走了len步,则fast走了2*len步,相遇时多走了一圈:
环长=2*len-len。
3.求有环单链表的环连接点位置
第一次碰撞点Pos到连接点Join的距离=头指针到连接点Join的距离,因此,分别从第一次碰撞点Pos、头指针head开始走,相遇的那个点就是连接点。

在环上相遇后,记录第一次相遇点为Pos,连接点为Join,假设头结点到连接点的长度为LenA,连接点到第一次相遇点的长度为x,环长为R。
第一次相遇时,slow走的长度 S = LenA + x;
第一次相遇时,fast走的长度 2S = LenA + n*R + x;
所以可以知道,LenA + x = n*R; LenA = n*R -x;
4.求有环单链表的链表长
上述2中求出了环的长度;3中求出了连接点的位置,就可以求出头结点到连接点的长度。两者相加就是链表的长度。
编程实现:
下面是代码中的例子:

具体代码如下:
#include <stdio.h> #include <stdlib.h> typedef struct node{ int value; struct node *next; }LinkNode,*Linklist; /// 创建链表(链表长度,环节点起始位置) Linklist createList(){ Linklist head = NULL; LinkNode *preNode = head; LinkNode *FifthNode = NULL; for(int i=0;i<6;i++){ LinkNode *tt = (LinkNode*)malloc(sizeof(LinkNode)); tt->value = i; tt->next = NULL; if(preNode == NULL){ head = tt; preNode = head; } else{ preNode->next =tt; preNode = tt; } if(i == 3) FifthNode = tt; } preNode->next = FifthNode; return head; } ///判断链表是否有环 LinkNode* judgeRing(Linklist list){ LinkNode *fast = list; LinkNode *slow = list; if(list == NULL) return NULL; while(true){ if(slow->next != NULL && fast->next != NULL && fast->next->next != NULL){ slow = slow->next; fast = fast->next->next; } else return NULL; if(fast == slow) return fast; } } ///获取链表环长 int getRingLength(LinkNode *meetNode){ int RingLength=0; LinkNode *fast = meetNode; LinkNode *slow = meetNode; for(;;){ fast = fast->next->next; slow = slow->next; RingLength++; if(fast == slow) break; } return RingLength; } ///获取链表头到环连接点的长度 int getLenA(Linklist list,LinkNode *meetNode){ int lenA=0; LinkNode *fast = list; LinkNode *slow = meetNode; for(;;){ fast = fast->next; slow = slow->next; lenA++; if(fast == slow) break; } return lenA; } ///释放空间 int freeMalloc(Linklist list){ LinkNode *nextnode = NULL; while(list != NULL){ nextnode = list->next; free(list); list = nextnode; } } int main(){ Linklist list = NULL; LinkNode *meetNode = NULL; int RingLength = 0; int LenA = 0; list = createList(); meetNode = judgeRing(list); if(meetNode == NULL) printf("No Ring\n"); else{ printf("Have Ring\n"); RingLength = getRingLength(meetNode); LenA = getLenA(list,meetNode); printf("RingLength:%d\n",RingLength); printf("LenA:%d\n",LenA); printf("listLength=%d\n",RingLength+LenA); freeMalloc(list); } return 0; }
1 #include <stdio.h>
2 #include <stdlib.h>
3 typedef struct node{
4 int value;
5 struct node *next;
6 }LinkNode,*Linklist;
7
8 /// 创建链表(链表长度,环节点起始位置)
9 Linklist createList(){
10 Linklist head = NULL;
11 LinkNode *preNode = head;
12 LinkNode *FifthNode = NULL;
13 for(int i=0;i<6;i++){
14 LinkNode *tt = (LinkNode*)malloc(sizeof(LinkNode));
15 tt->value = i;
16 tt->next = NULL;
17 if(preNode == NULL){
18 head = tt;
19 preNode = head;
20 }
21 else{
22 preNode->next =tt;
23 preNode = tt;
24 }
25
26 if(i == 3)
27 FifthNode = tt;
28 }
29 preNode->next = FifthNode;
30 return head;
31 }
32
33 ///判断链表是否有环
34 LinkNode* judgeRing(Linklist list){
35 LinkNode *fast = list;
36 LinkNode *slow = list;
37
38 if(list == NULL)
39 return NULL;
40
41 while(true){
42 if(slow->next != NULL && fast->next != NULL && fast->next->next != NULL){
43 slow = slow->next;
44 fast = fast->next->next;
45 }
46 else
47 return NULL;
48
49 if(fast == slow)
50 return fast;
51 }
52 }
53
54 ///获取链表环长
55 int getRingLength(LinkNode *meetNode){
56 int RingLength=0;
57 LinkNode *fast = meetNode;
58 LinkNode *slow = meetNode;
59 for(;;){
60 fast = fast->next->next;
61 slow = slow->next;
62 RingLength++;
63 if(fast == slow)
64 break;
65 }
66 return RingLength;
67 }
68
69 ///获取链表头到环连接点的长度
70 int getLenA(Linklist list,LinkNode *meetNode){
71 int lenA=0;
72 LinkNode *fast = list;
73 LinkNode *slow = meetNode;
74 for(;;){
75 fast = fast->next;
76 slow = slow->next;
77 lenA++;
78 if(fast == slow)
79 break;
80 }
81 return lenA;
82 }
83
84 ///释放空间
85 int freeMalloc(Linklist list){
86 LinkNode *nextnode = NULL;
87 while(list != NULL){
88 nextnode = list->next;
89 free(list);
90 list = nextnode;
91 }
92 }
93
94 int main(){
95 Linklist list = NULL;
96 LinkNode *meetNode = NULL;
97 int RingLength = 0;
98 int LenA = 0;
99
100 list = createList();
101 meetNode = judgeRing(list);
102
103 if(meetNode == NULL)
104 printf("No Ring\n");
105 else{
106 printf("Have Ring\n");
107 RingLength = getRingLength(meetNode);
108 LenA = getLenA(list,meetNode);
109
110 printf("RingLength:%d\n",RingLength);
111 printf("LenA:%d\n",LenA);
112 printf("listLength=%d\n",RingLength+LenA);
113
114 freeMalloc(list);
115 }
116 return 0;
117 }
执行结果:

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原文地址:http://www.cnblogs.com/bendantuohai/p/4499356.html