160.Intersection-of-Two-Linked-Lists

160. Intersection of Two Linked Lists

题目地址

题目描述

Write a program to find the node at which the intersection of two singly linked lists begins.
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For example, the following two linked lists:
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begin to intersect at node c1.
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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.
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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.
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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.
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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.

代码

Approach #1 Hash Table

/**
* 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) {
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}
}

Approach #2 Two Pointers

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

Approach #2

loop the other list
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
// boundary check
if (headA == null || headB == null) return null;
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ListNode a = headA;
ListNode b = headB;
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// 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;
}
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return a;
}