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:
null.这道题还是先要搞懂题目问的啥。这里的intersection linked list通过看例子我们可以看出，其length可以不一样，但是从intersect的部分开始到结束长度是一样的，所以说多余的只会是前面的几个值，而且一定有较长length的那个linked list多余出来的那一段的值。
所以我们首先计算两个list的length，把较长的那个的多余的那部分的值给waive掉。这样两条同样长的list就非常好找intersect的地方了。
代码如下。~
/**
* 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 lengtha=0;
int lengthb=0;
ListNode curr1=headA;
ListNode curr2=headB;
while(curr1!=null){
lengtha++;
curr1=curr1.next;
}
while(curr2!=null){
lengthb++;
curr2=curr2.next;
}
curr1=headA;
curr2=headB;
if(lengtha>lengthb){
for(int i=0;i<lengtha-lengthb;i++){
curr1=curr1.next;
}
}else{
for(int i=0;i<lengthb-lengtha;i++){
curr2=curr2.next;
}
}
while(curr1!=curr2){
curr1=curr1.next;
curr2=curr2.next;
}
return curr1;
}
}