一、判断两个单链表是否是相交
ide
思路分析:对象
最简单直接的方法就是依次遍历两条链表,判断其尾节点是否是相等,相等则相交不然不相交。it
bool CheckCross(const List& list1, const List& list2)//list1,list2为两个对象 { Node* l1 = list1._head; Node* l2 = list2._head; while (l1->_next)//找到list1的尾节点 { l1 = l1->_next; } while (l2->_next)//找到list2的尾节点 { l2 = l2->_next; } if (l1 == l2) { return true;//相交 } return false;//不相交 }
二、找到两个单链表的交点ast
思路分析:class
在两个单链表长度相等的状况下是最简单的,只须要同时遍历两个链表而且不断地比较,若是相等则为交点不然不是交点。可是在两条单链表长度不相等的状况下,则能够让长度较长的链表先遍历两条链表的长度之差,而后再同时遍历既可。List
Node* GetCrossNode(List& list1, List& list2) { int length1 = list1.GetListLength();//求list1的长度 int length2 = list2.GetListLength();//求list2的长度 int diff = 0; Node* slow = NULL; Node* fast = NULL; if (length1 > length2)//list1的长度比list2的长度要长 { diff = length1 - length2;//单链表的长度之差 fast = list1._head; slow = list2._head; while (diff--) { fast = fast->_next; } while (fast&&slow) { if (fast == slow) { return slow;//返回交点 } fast = fast->_next; slow = slow->_next; } } else //list2的长度比list1的长度要长 { diff = length1 - length2; fast = list2._head; slow = list1._head; while (diff--) { fast = fast->_next; } while (fast&&slow) { if (fast == slow) { return slow;//返回交点 } fast = fast->_next; slow = slow->_next; } } return NULL;//不相交 } int List::GetListLength()//求取单链表长度的方法 { int count = 0; Node* cur = _head; while (cur) { count++; cur = cur->_next; } return count; }