#leetcode#Intersection of Two LinkedList

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.

分析： 找两个linkedlist相交节点，首先得到两个LinkedList A， B 的长度，求得A，B的长度差diff，维护two pointers，长的LinkedList 的 pointer先移动 diff 步，然后两个pointer同时移动，则如果相交的花两个pointer必然移动到同一结点

/**
 * 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) {
        if(headA == null || headB == null){
            return null;
        }

        int lenA = getLength(headA);
        int lenB = getLength(headB);
        ListNode shortHead = lenA > lenB ? headB : headA;
        ListNode longHead = lenA > lenB ? headA : headB;
        int diff = Math.abs(lenA - lenB);
        for(int i = 0; i < diff; i++){
            longHead = longHead.next;
        }
        while(shortHead != null){
            if(shortHead == longHead){
                return shortHead;
            }
            shortHead = shortHead.next;
            longHead = longHead.next;
        }

        return null;
    }

    private int getLength(ListNode head){
        int res = 0;
        while(head != null){
            res++;
            head = head.next;
        }
        return res;
    }
}

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