题目
给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表不存在相交节点，返回 null 。
输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
输出：Intersected at '8'
解题思路：
我们将每个链表分为两部分，不相交部分和相交部分C
那么listA = A+C listB = B+C 他们的总长度为：A+C+B = B+C+A
所以我们只要找出两个节点相等的地方就是交点,因为他们只要相交的话一定会相遇的
逐步演示双指针过程
假设listA = [4,1,8] listB = [5,6,1,8]
初始状态`
`pA -> 4`
`pB -> 5`
`第1步：`
`pA = 4, pB = 5，不相等`
`pA = pA.next = 1`
`pB = pB.next = 6`
`第2步`
`pA = 1, pB = 6，不相等`
`pA = pA.next = 8`
`pB = pB.next = 1`
`第3步`
`pA = 8, pB = 1，不相等`
`pA = pA.next = None (链表A结束)`
`pB = pB.next = 8`
`第4步`
`pA = None, pB = 8，不相等`
`pA = headB = 5 (切换到链表B)`
`pB = pB.next = None`
`第5步`
`pA = 5, pB = None，不相等`
`pA = pA.next = 6`
`pB = headA = 4 (切换到链表A)`
`第6步`
`pA = 6, pB = 4，不相等`
`pA = pA.next = 1`
`pB = pB.next = 1`
`第7步`
`pA = 1, pB = 1，相等 ✓ 找到相交节点！`
`// 定义链表节点类`
`class ListNode {`
` constructor(val) {`
` this.val = val;`
` this.next = null;`
` }`
`}`
`function getIntersectionNode (headA,headB){`
` if(!headA || !headB) return null;`
` let nodeA = headA`
` let nodeB = headB`
` while(nodeA !== nodeB ){`
` nodeA = nodeA !== null ? nodeA.next : headB`
` nodeB = nodeB !== null ? nodeB.next : headA`
` }`
` return nodeA`
`}`
`// 示例：手动创建链表并传入`
`// 创建 listA: 4 -> 1 -> 8 -> 4 -> 5`
`let a1 = new ListNode(4);`
`let a2 = new ListNode(1);`
`let a3 = new ListNode(8);`
`let a4 = new ListNode(4);`
`let a5 = new ListNode(5);`
`a1.next = a2;`
`a2.next = a3;`
`a3.next = a4;`
`a4.next = a5;`
`// 创建 listB: 5 -> 6 -> 1 -> 8 -> 4 -> 5`
`let b1 = new ListNode(5);`
`let b2 = new ListNode(6);`
`let b3 = new ListNode(1);`
`let b4 = new ListNode(8);`
`let b5 = new ListNode(4);`
`let b6 = new ListNode(5);`
`b1.next = b2;`
`b2.next = b3;`
`b3.next = b4;`
`b4.next = b5;`
`b5.next = b6;`
`// 设置相交点：listA 的第3个节点（8）与 listB 的第4个节点（8）相交`
`// 实际上是 listB 的第4个节点指向 listA 的第3个节点`
`b4.next = a3; // b4 (8) -> a3 (8)`
`let headA = a1;`
`let headB = b1;`
`console.log(getIntersectionNode(headA, headB)?.val); // 输出 8`
`