160. Intersection of Two Linked Lists

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转载最新推荐文章于 2022-08-06 10:20:24 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/jessie2009/p/9817727.html

本文介绍了一种在两个单链表中找到它们开始相交节点的算法。通过首先确定两个链表的长度差异，然后同步移动两个指针，直到它们相遇在交点处或同时到达链表尾部。该算法在O(n)时间内运行，仅使用O(1)额外空间。

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.

//Time: O(n), Space: O(1)    
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        
        int lenA = getLen(headA);
        int lenB = getLen(headB);
        
        if (lenA > lenB) {
            headA = forward(headA, lenA - lenB);
        } else {
            headB = forward(headB, lenB - lenA);
        }
        
        
        while (headA != null && headB != null) {
            if (headA == headB) {
                return headA;
            }
            
            headA = headA.next;
            headB = headB.next;
        }
        
        return null;
    }
    
    private int getLen(ListNode head) {
        int count = 1;
        
        while (head != null) {
            head = head.next;
            count++;
        }
        
        return count;
    }
    
    private ListNode forward(ListNode head, int step) {
        while (step > 0) {
            head = head.next;
            step--;
        }
        
        return head;
    }

转载于:https://www.cnblogs.com/jessie2009/p/9817727.html