160. 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.

含义：两个列表有交集，求出他们的交集开始点

方法一：

    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode a = headA;
        ListNode b = headB;
        //if a & b have different len, then we will stop the loop after second iteration
        while( a != b){
            //for the end of first iteration, we just reset the pointer to the head of another linkedlist
            a = a == null? headB : a.next;
            b = b == null? headA : b.next;
        }
        return a; //如果a不等于null，代表找到了交集的开始点  如果a等于null，说明两个列表没有交集

}

方法二：

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

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int lengthA = getListLength(headA);
        int lengthB = getListLength(headB);
        while (lengthA > lengthB) {
            lengthA--;
            headA = headA.next;
        }
        while (lengthB > lengthA) {
            lengthB--;
            headB = headB.next;
        }
        while (headA != headB) {
            headA = headA.next;
            headB = headB.next;
        }
        return headA; //如果headA不等于null，代表找到了交集的开始点  如果headA等于null，说明两个列表没有交集

}

 

方法三：解决方法就是将A链表尾节点指向头结点形成一个环，检测B链表是否存在环，如果存在，则两个链表相交，而检测出来的依赖环入口即为相交的第一个点

    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == headB || headA==null) return headA;
        ListNode temp = headA;
        while (temp.next!=null) temp = temp.next;
        temp.next = headA;
        ListNode slow = headB,fast = headB;
        while (fast!=null && fast.next !=null)
        {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast)
            {
                while (headB != slow)
                {
                    headB = headB.next;
                    slow = slow.next;
                }
                return slow;
            }

        }
        return null;
    }