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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：

 1 class Solution {
 2 public:
 3     /**
 4      * @param headA: the first list
 5      * @param headB: the second list
 6      * @return: a ListNode
 7      */
 8     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
 9         // write your code here
10         if(headA == NULL || headB == NULL)
11             return NULL;
12         ListNode* iter1 = headA;
13         ListNode* iter2 = headB;
14         int len1 = 1;
15         while(iter1->next != NULL)
16         {
17             iter1 = iter1->next;
18             len1 ++;
19         }
20         int len2 = 1;
21         while(iter2->next != NULL)
22         {
23             iter2 = iter2->next;
24             len2 ++;
25         }
26         if(iter1 != iter2)
27             return NULL;
28         if(len1 > len2)
29         {
30             for(int i = 0; i < len1-len2; i ++)
31                 headA = headA->next;
32         }
33         else if(len2 > len1)
34         {
35             for(int i = 0; i < len2-len1; i ++)
36                 headB = headB->next;
37         }
38         while(headA != headB)
39         {
40             headA = headA->next;
41             headB = headB->next;
42         }
43         return headA;
44     }
45 };

先算长度，然后长的先走差值步，然后同时走

 

解法二：

 1 public class Solution {
 2     /**
 3      * @param headA: the first list
 4      * @param headB: the second list
 5      * @return: a ListNode 
 6      */
 7     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 8         if (headA == null || headB == null) {
 9             return null;
10         }
11         
12         // get the tail of list A.
13         ListNode node = headA;
14         while (node.next != null) {
15             node = node.next;
16         }
17         node.next = headB;
18         ListNode result = listCycleII(headA);
19         node.next = null;
20         return result;
21     }
22     
23     private ListNode listCycleII(ListNode head) {
24         ListNode slow = head, fast = head.next;
25         
26         while (slow != fast) {
27             if (fast == null || fast.next == null) {
28                 return null;
29             }
30             
31             slow = slow.next;
32             fast = fast.next.next;
33         }
34         
35         slow = head;
36         fast = fast.next;
37         while (slow != fast) {
38             slow = slow.next;
39             fast = fast.next;
40         }
41         
42         return slow;
43     }
44 }

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】）

posted on 2017-10-14 18:58  LastBattle  阅读(205)  评论(0)  编辑  收藏  举报