算法 - 检测是否链表存在环

 

 

这儿例子使用快慢指针的办法实现。

 

(H)      LengthX      (E)

------------------------+------------+

                               |               |

                               |               |

                               |               |  (M)  (E->M: LengthY)

                               +-----------+

                     (M->E: LengthZ)

 

Length of circle = LengthY + LengthZ

 

 

 

假定

1. 开始点是H, 环的进入点是E, 快指针和慢指针首次相遇的点是M.

2. 慢指针和快指针首次相遇在M点的时候, 是慢指针首次到达M点, 而此时快指针是第N(N>=2)次到达M点。

3. 此后如何快指针慢指针还是按照原来的逻辑走, 那么当下一次慢指针到达M点的时候, 快指针和慢指针又相遇与M点, 也就是说以后每次他们俩都相遇与M点, 而此时慢指针正好又走了一圈, 而快指针正好又走了两圈。

 

 

 

-- 计算入口点E和环长度--

当慢指针首次到达M点的时候, 慢指针走过的路程是(H->E) 和(E->M)其长度为LengthX+LengthY

而此时快指针走过的长度为2(LengthX+LengthY), 因为慢指针都一步, 快指针走两步.

 

快指针走过的长度 = LengthX + LengthY + (LengthY + LengthZ) * N = 2 * (LengthX+LengthY)   // N >= 1

                        => (LengthY + LengthZ) * N = (LengthX+LengthY)

 

 

当把快慢指针分别从M点按照步进1的速度往回退的时候, 当慢指针回退到H点时, 快指针正好回退到M点, 此时把快慢指针分别按照步进1的速度前进, 当他们相遇的时候就是环的第一个入口点E。

 

Code start

struct tnode{ int data; // node content struct tnode *next; int visited; // flag whether this node has been visited or not};typedef struct tnode node;node *pf=head;node *ps=head;while(pf &&pf->next){ pf = pf->next->next; ps = ps->next; if(pf == ps) { printf("Yes, I have a circle, they met at "); printNode(ps); break; }}if (pf == NULL){ printf("I have no circle./n"); return;}// circle again to make sure another circle will meet on same node.pf = ps->next->next;ps = ps->next;while(pf != ps){ pf = pf->next->next; ps = ps->next;}printf("Yes, I have a circle, circle againg, they met at ");printNode(ps);//calculate the length of the circle.int length = 0;do{ ps = ps->next; pf = pf->next->next; length++;} while ( ps != pf );printf("Yes, I have a circle, circle size=%d, they met at ",length);printNode(ps);//find the entry node of circle.pf=head;while(pf!=ps){ pf=pf->next; ps=ps->next;}printf("Yes, I have a circle, entry node is ");printNode(ps);

Code end