*U'`CJ6c)R《编程之美》3.6节“链表相交”扩展问题答案BSD爱好者乐园 PttIK-a"S
?q
扩展1:链表1 步长为1, 链表2步长为2 ,如果有环且相交则肯定相遇,否则不相交
Un;?3Sm"G!E'M(Wilist1 head: p1
6Vc jj!P;i-\Dv
R.U5Rlist2 head: p2BSD爱好者乐园2[Sg]k"B"O5R/Yb
while( p1 != p2 && p1 != NULL && p2 != NULL )
F5SvIKh8H
v{
pn2X,?HW5| ] p1 = p1->next;
4ES3H:WRd if ( p2->next )
J%x%dst gf(x p2 = p2->next->next;
S ]IS1~u8f$y else