leetcode 160 Intersection of Two Linked Lists
z
 2018-03-01 20:36:24  

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

For example, the following two linked lists:

1
2
3
4
5
6
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.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
/**
 * 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;
        int lengthA = 0;
        int lengthB = 0;
        int start = 0;
        while(a!=null){
            a = a.next;
            lengthA ++;
        }
        while(b!=null){
            b = b.next;
            lengthB ++;
        }
        if(lengthA > lengthB){
            b = headB;
            a = headA;
            int i = 0;
            while(i < lengthA - lengthB){
                a = a.next;
                i ++;
            }
        }
        if(lengthB >= lengthA){
            a = headA;
            b = headB;
            int i = 0;
            while(i < lengthB - lengthA){
                b = b.next;
                i++;
            }
        }
        while(a != null && b != null){
            if(a.val != b.val){
                a = a.next;
                b = b.next;
            }
            else{
                return a;
            }
        }
        return null;
        
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    //boundary check
    if(headA == null || headB == null) return null;
    
    ListNode a = headA;
    ListNode b = headB;
    
    //if a & b have different len, then we will stop the loop after second iteration
    while( a != b){
    	//for the end of first iteration, we just reset the pointer to the head of another linkedlist
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }
    
    return a;
}