160. Intersection of Two Linked Lists【easy】

160. Intersection of Two Linked Lists【easy】

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

解法一：

 class Solution {
 public:
     /**
      * @param headA: the first list
      * @param headB: the second list
      * @return: a ListNode
      */
     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
         // write your code here
         if(headA == NULL || headB == NULL)
             return NULL;
         ListNode* iter1 = headA;
         ListNode* iter2 = headB;
         int len1 = ;
         while(iter1->next != NULL)
         {
             iter1 = iter1->next;
             len1 ++;
         }
         int len2 = ;
         while(iter2->next != NULL)
         {
             iter2 = iter2->next;
             len2 ++;
         }
         if(iter1 != iter2)
             return NULL;
         if(len1 > len2)
         {
             for(int i = ; i < len1-len2; i ++)
                 headA = headA->next;
         }
         else if(len2 > len1)
         {
             for(int i = ; i < len2-len1; i ++)
                 headB = headB->next;
         }
         while(headA != headB)
         {
             headA = headA->next;
             headB = headB->next;
         }
         return headA;
     }
 };

先算长度，然后长的先走差值步，然后同时走

解法二：

 public class Solution {
     /**
      * @param headA: the first list
      * @param headB: the second list
      * @return: a ListNode
      */
     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
         if (headA == null || headB == null) {
             return null;
         }

         // get the tail of list A.
         ListNode node = headA;
         while (node.next != null) {
             node = node.next;
         }
         node.next = headB;
         ListNode result = listCycleII(headA);
         node.next = null;
         return result;
     }

     private ListNode listCycleII(ListNode head) {
         ListNode slow = head, fast = head.next;

         while (slow != fast) {
             if (fast == null || fast.next == null) {
                 return null;
             }

             slow = slow.next;
             fast = fast.next.next;
         }

         slow = head;
         fast = fast.next;
         while (slow != fast) {
             slow = slow.next;
             fast = fast.next;
         }

         return slow;
     }
 }

先弄成环，转换为找环的入口问题，找到之后再断开环

找环的问题解法可以参见（142. Linked List Cycle II【easy】）