LeetCode Intersection of Two Linked Lists

原题链接在这里：https://leetcode.com/problems/intersection-of-two-linked-lists/

题目：

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.

题解：

找到length diff, 长的list head先移动diff次, 再一起移动找相同点.

Time Complexity: O(len1 + len2), len1 is the length of list one. len2 is the length of list two.

Space: O(1).

AC Java:

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
         int len1 = length(headA);
         int len2 = length(headB);
         while(len1 > len2){
             headA = headA.next;
             len1--;
         }

         while(len2 > len1){
             headB =headB.next;
             len2--;
         }

         while(headA != headB){
             headA = headA.next;
             headB = headB.next;
         }

         return headA;
     }

     private int length(ListNode head){
         int len = 0;
         while(head != null){
             head = head.next;
             len++;
         }
         return len;
     }
 }

有一巧妙地方法来综合掉 length diff, a = headA, b = headB, a和b一起移动。当a到了list A的末位就跳到HeadB, b到了List B的末位就跳到HeadA.

等a和b相遇就是first intersection node. 因为a把first intersection node之前的list A部分, list B部分都走了一次. b也是如此. diff就综合掉了.

若是没有intersection, 那么a走到list B的结尾 null时, b正好走到 list A的结尾null, a==b. 返回了null.

Time Complexity: O(len1 + len2), len1 is the length of list one. len2 is the length of list two.

Space: O(1).

AC  Java:

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
         ListNode a = headA;
         ListNode b = headB;
         while(a != b){
             a = a==null ? headB : a.next;
             b = b==null ? headA : b.next;
         }
         return a;
     }
 }