160 Intersection of Two Linked List

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.

Method1 : Find the length of two linked list and then move one list to same length

/**
 * 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 = len(headA);
        int lenB = len(headB);
        ListNode curtA = headA;
        ListNode curtB = headB;
        if (lenA != lenB) {
            if (lenA > lenB) {
                for (int i = 0; i < lenA - lenB; i++) {
                    curtA = curtA.next;
                }
            } else if (lenB > lenA){
                for (int i = 0; i < lenB - lenA; i++) {
                    curtB = curtB.next;
                }
            }
        }

        while (curtA != null && curtB != null) {
            if (curtA.val!=curtB.val) {
                curtA  = curtA.next;
                curtB = curtB.next;
            }else{
                return curtA;
            }
        }
        return null;
    }
    public int len(ListNode head) {
        int length = 0;
        while (head != null) {
            head = head.next;
            length++;
        }
        return length;
    }
}

另一种方法很巧妙时间复杂度为O(M+n)

 /**
     *
     A:          a1 → a2
                         ↘
                         c1 → c2 → c3
                         ↗
     B:     b1 → b2 → b3
     begin to intersect at node c1.
     A : a1 → a2 -> c1 → c2 → c3 -> b1 → b2 → b3 -> c1 → c2 → c3
     B : b1 → b2 → b3 -> c1 → c2 → c3 -> a1 → a2 -> c1 → c2 → c3
     time : O(m + n);
     space : O(1);
     * @param headA
     * @param headB
     * @return
     */

    public ListNode getIntersectionNode2(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) return null;
        ListNode a = headA;
        ListNode b = headB;
        while (a != b) {
            a = a == null ? headB : a.next;
            b = b == null ? headA : b.next;
        }
        return a;
    }
}