LeetCode 刷题【160. 相交链表】

160. 相交链表

自己做

解：快慢指针

/**
 * 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 len1 = 0;
        int len2 = 0;
        ListNode p = headA;
        ListNode q = headB;
        
        while(p != null){
            len1++;
            p = p.next;
        }

        while(q != null){
            len2++;
            q = q.next;
        }

        p = headA;
        q = headB;

        if(len1 < len2)
            for(int i = 0; i < len2 - len1; i++)
                q = q.next;
        else
            for(int i = 0; i < len1 - len2; i++)
                p = p.next;

        while(p != null && q != null && p != q){
            p = p.next;
            q = q.next;
        }        

        if(p == q)
            return p;
        else
            return null;
    }
}