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LeetCode Top 100 Liked Questions 160. Intersection of Two Linked Lists (Java版; Easy)
题目描述
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
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.
第一次做
/*
找出两个无环链表的交点
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if((headA==null&&headB!=null) || (headA!=null&&headB==null))
return null;
if(headA==null && headB==null)
return null;
int n = 0;
ListNode curr = headA;
while(curr!=null){
n++;
curr = curr.next;
}
curr = headB;
while(curr!=null){
n--;
curr = curr.next;
}
if(n>0){
while(n>0){
headA = headA.next;
n--;
}
}
else{
n = Math.abs(n);
while(n>0){
headB = headB.next;
n--;
}
}
while(headA!=null && headB!=null){
if(headA==headB)
return headA;
headA = headA.next;
headB = headB.next;
}
return null;
}
}
