Leetcode 160 Intersection of Two Linked Lists(快慢指针)

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.

题目解析：

Leetcode 里面好多链表的题都可以用快慢指针来解. 

这道题的思路就是先遍历两个lists然后得到两个lists的长度, 然后让长的那个lists先走Math.abs(lengthA-lengthB)步, 然后一起走判断下一个节点是不是相等.

注意其中有一步就是两个lists遍历之后要判断最后一个节点是不是相等, 不等就可以直接返回null啦~

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    if(headA == null && headB == null)
        return null;
    ListNode node1 = headA;
    ListNode node2 = headB;
    
    int lengthA = 1;
    int lengthB = 1;
    while(node1 != null){
        lengthA++;
        node1 = node1.next;
    }
    while(node2 != null){
        lengthB++;
        node2 = node2.next;
    }

    if(node1 != node2)    //先判断一下
        return null;
    else{
        int count = Math.abs(lengthA - lengthB);
        if(lengthA > lengthB){    //此处处理要注意
            node1 = headA;
            node2 = headB;
        }else{
            node2 = headA;
            node1 = headB;
        }
        for(int i = 0; i < count; i++){
            node1 = node1.next;
        }
        while(node1 != null && node2 != null & node1 != node2){
            node1 = node1.next;
            node2 = node2.next;
        }
        
    }
    return node1;
}