面试题 02.07. 链表相交-双指针法

/**
 * 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) {
        int lenA = 0;
        int lenB = 0;
        ListNode curA = headA;
        ListNode curB = headB;
        //获取链表A的长度
        while(curA!=null){
            lenA++;
            curA=curA.next;
        }
        //获取链表B的长度
         while(curB!=null){
            lenB++;
            curB=curB.next;
        }
        curA = headA;
        curB = headB;
        //让curA为最长链表的头，lenA为其长度
        if(lenA<lenB){
            //swap(lenA,lenB);
            int temp=lenA;
            lenA=lenB;
            lenB=temp;
            //swap(curA,curB);
            ListNode tempNode = curA;
            curA=curB;
            curB=tempNode;
        }
        //求出两个链表长度的差值
        int gap = lenA-lenB;
        //让长的链表与端的链表尾部对齐
        while(gap-- >0){
            curA = curA.next;
        }
        //比较curA和curB是否相同
        while(curA!=null){
            if(curA == curB){ //相同就结束返回
                return curA;
            }
            //不相同，同时向后移动curA和curB
            curA=curA.next;
            curB=curB.next;
        }
        return null;
    }
}
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