[LeetCode] Intersection of Two Linked Lists

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.

     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;
    }
}

 

转载于:https://www.cnblogs.com/orangeme404/p/4747667.html