烟_火
人间烟火何不尝

编写一个程序,找到两个单链表相交的起始节点。

如下面的两个链表:

在节点 c1 开始相交

/**
 * 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) {
        ListNode NodeA = headA , NodeB = headB, returnNode = null;
        int numA = 0, numB = 0, tmp = 0;
        while(NodeA != null)
        {
            numA++;
            NodeA = NodeA.next;
        }
        while(NodeB != null)
        {
            numB++;
            NodeB = NodeB.next;
        }
        if(numA > numB)
        {
            tmp = numA - numB;
            while(tmp > 0)
            {
                headA = headA.next;
                tmp--;
            }
            while(headA != null)
            {
                if(headA == headB)
                {
                    returnNode = headA;
                    return returnNode;
                }
                headA = headA.next;
                headB = headB.next;
            }
        }
        else
        {
            tmp = numB - numA;
            while(tmp > 0)
            {
                headB = headB.next;
                tmp--;
            }
            while(headB != null)
            {
                
                if(headB == headA)
                {
                    returnNode = headB;
                    return returnNode;
                }
                headB = headB.next;
                headA = headA.next;
            }
        }
        return returnNode; 
    }
}
posted on 2020-03-25 21:38  段流儿  阅读(108)  评论(0编辑  收藏  举报