Question:

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

Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

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.

Thinking:

  • Method1:
    • 如果两条链表有交集。则a跑到尾开始跑B，b跑完后开始跑a。
    • 测试两个指针跑动的长度是相同的。
/**
 * 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) {
        if(headA == null || headB == null) return null;
        ListNode a = headA;
        ListNode b = headB;
        int count = 0;
        while(a != b){
            if(count == 2) break;
            if(a == null) count ++;
            a = a == null ? headB : a.next;
            b = b == null ? headA : b.next;
        }
        if(a != b) return null;
        else
            return a;
    }
}

二刷

这道题印象还是比较深的，就是没想到和一刷一样优雅的表达方式。后来还是参考了一刷的表达方式。

/**
 * 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) {
        if(headA == null || headB == null) return null;
        ListNode curA = headA;
        ListNode curB = headB;
        int count = 0;
        while(curA != curB){
            if(count == 2) break;
            if(curA == null) count++;
            curA = curA == null ? headB : curA.next;
            curB = curB == null ? headA : curB.next;
        }
        return (curA == curB) ? curA : null;
    }
}