判断两个无环链表是否相交
题目：给定两个无环的单链表，判断两者是否相交（即有共同的节点）。
思路：可以将其中一个链表连接到另一个链表尾部，并通过判定得到新的链表是否有环来确定两个链表是否相交。
代码：
1 #include <cstdio> 2 3 typedef struct node 4 { 5 struct node* next; 6 }Node, *pNode; 7 8 int bListIntersect (Node list1, Node list2); 9 10 int main (int argc, char** argv) 11 { 12 Node n1, n2, n3, n4, n5, n6, n7, n8, n9, n10; 13 n1.next = &n2; 14 n2.next = &n3; 15 n3.next = &n4; 16 n4.next = NULL; 17 18 n5.next = &n6; 19 n6.next = &n7; 20 n7.next = NULL; 21 22 n8.next = &n9; 23 n9.next = &n10; 24 n10.next = &n3; 25 26 printf(bListIntersect(n1, n5) == 1?"intersect\n":"not intersect\n"); 27 printf(bListIntersect(n1, n8) == 1?"intersect\n":"not intersect\n"); 28 29 return 0; 30 } 31 32 int bListIntersect (Node list1, Node list2) 33 { 34 pNode p, q; 35 p = &list1; 36 while (p->next != NULL) 37 { 38 p = p->next; 39 } 40 41 p->next = &list2; 42 q = p; 43 p = p->next; 44 while ((p != NULL) && (p != q)) 45 { 46 p = p->next; 47 } 48 q->next = NULL; 49 return (p == NULL) ? 0 : 1; 50 }
