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  • 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.
    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) {
     *         val = x;
     *         next = null;
     *     }
     * }
     */
    public class Solution {
        
        public int getListLength(ListNode N){
            int n = 0;
            ListNode p = N;
            while(p != null){
                n++;
                p = p.next;
            }
            return n;
        }
        
        public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
          
            ListNode intersect = null;
            ListNode pA = headA;
            ListNode pB = headB;
            int A = getListLength(headA);
            int B = getListLength(headB);
            if(A >= B){
                int m = A-B;
                while(m != 0){
                    pA = pA.next;
                    m--;
                }
            }
            else{
                int m = B-A;
                while(m != 0){
                    pB = pB.next;
                    m--;
                }
            }
            
            while(pA != null){
                if(pA == pB){
                    intersect = pA;
                    break;
                }
                else{
                    pA = pA.next;
                    pB = pB.next;
                }
            }
            
            return intersect;
        }
    }

    思路很简单,长的链表先移动指针至和短的链表“同一起跑线”,然后同时移动指针并比较是否相同。

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  • 原文地址:https://www.cnblogs.com/mrpod2g/p/4250709.html
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