LinkedList-160-Intersection of Two Linked Lists

Description:

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.

Solution:

//依题意知.A与B的intersection表示从相交点到尾端都重合，因此取A和B长度较大者的头指针右移直到长度较大者的剩余节点长度
//与较小者相等(只有这样才可能相交),然后遍历，headA与headB相等处即为交点
/**
* 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 lenA = length(headA), lenB = length(headB);
// move headA and headB to the same start point
while (lenA > lenB) {
headA = headA.next;
lenA--;
}
while (lenA < lenB) {
headB = headB.next;
lenB--;
}
// find the intersection until end
while (headA != headB) {
headA = headA.next;
headB = headB.next;
}
return headA;
}

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