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链表题型可分为快慢指针和虚拟头节点两种解题技巧。
快慢指针
使用两个指针（快指针和慢指针），以不同的速度遍历链表，解决与链表位置、环检测相关的问题。
反转链表
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快慢指针，慢指针一次走一步，快指针一次走两步，直到快指针为null，返回慢指针节点
public ListNode middleNode(ListNode head){ListNode slow = head;ListNode fast = head;while (fast != null && fast.next != null){slow = slow.next;fast = fast.next.next;}return slow;}判断回文链表
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找到链表中间节点 + 反转后段链表，然后比较前后两段链表是否相等
环形链表
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判断是否有环：快慢指针，慢指针一次走一步，快指针一次走两步，如果相遇说明有环
public boolean hasCycle(ListNode head) {ListNode last = head;ListNode fast = head;while (fast != null && fast.next != null){if (fast == last){return true;}fast = fast.next.next;last = last.next;}return false;}-
找到环的首个节点：first指针指向头节点，second指针指向当前快指针，两个指针同时遍历到相遇
if (last == fast){ListNode first = head;ListNode second = fast;while (first != second){first = first.next;second = second.next;}return first;
}删除链表倒数第N个节点
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快慢指针，快指针先走n个节点，然后快慢指针同时移动，当快指针为null，删除慢指针。
public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dumy = new ListNode(0);dumy.next = head;ListNode last = dumy;ListNode fast = head;for (int i = 0; i < n; i ++){fast = fast.next;}while (fast != null){fast = fast.next;last = last.next;}last.next = last.next.next;return dumy.next;}虚拟头节点
在链表头部添加一个虚拟节点（dummy node），简化边界条件处理，尤其是涉及头节点操作的问题。
合并两个升序链表
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创造虚拟头节点，然后比较两个链表，依次向后添加
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode dump = new ListNode(0);ListNode cur = dump;while (list1 != null && list2 != null){if (list1.val < list2.val){cur.next = list1;list1 = list1.next;}else{cur.next = list2;list2 = list2.next;}cur = cur.next;}if (list1 != null){cur.next = list1;}if (list2 != null){cur.next = list2;}return dump.next;}链表相加
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创造虚拟头节点，依次相加两链表节点，sum = x + y + carry，carry = sum/10，sum = sum % 10
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode dump = new ListNode(0);ListNode cur = dump;int carry = 0;while (l1 != null || l2 != null){int x = 0, y = 0;if (l1 != null){x = l1.val;}if (l2 != null){y = l2.val;}int sum = x + y + carry;carry = sum / 10;sum = sum % 10;cur.next = new ListNode(sum);cur = cur.next;if (l1 != null){l1 = l1.next;}if (l2 != null){l2 = l2.next;}}if (carry == 1){cur.next = new ListNode(carry);}return dump.next;}两两交换链表中的节点
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创造虚拟头节点，创建one节点指向虚拟头节点，two节点 = one.next，three节点 = one.next.next，temp提前存one.next.next.next，然后one.next = three，three.next = two，two.next = temp
public ListNode swapPairs(ListNode head) {ListNode dumy = new ListNode(0);dumy.next = head;ListNode one = dumy;ListNode two;ListNode three;while (one.next != null && one.next.next != null){ListNode temp = one.next.next.next;two = one.next;three = one.next.next;one.next = three;three.next = two;two.next = temp;one = two;}return dumy.next;}排序链表
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归并排序，把当前链表一分为二（找到链表中间节点），再对左右子链表递归排序（合并两个升序链表）
public ListNode sortList(ListNode head) {if (head == null || head.next == null){return head;}ListNode mid = findMiddle(head);ListNode rightHead = mid.next;mid.next = null;ListNode left = sortList(head);ListNode right = sortList(rightHead);return mergeTwoLists(left, right);}private ListNode findMiddle(ListNode head){ListNode slow = head;ListNode fast = head.next;while (fast != null && fast.next != null){slow = slow.next;fast = fast.next;}return slow;}private ListNode mergeTwoLists(ListNode l1, ListNode l2){ListNode dummy = new ListNode(0);ListNode curr = dummy;while (l1 != null && l2 != null){if (l1.val < l2.val){curr.next = l1;l1 = l1.next;}else {curr.next = l2;l2 = l2.next;}curr = curr.next;}if (l1 != null){curr.next = l1;}if (l2 != null){curr.next = l2;}return dummy.next;}其他
相交链表
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两个指针，先遍历出两个链表的长度，然后长链表和短链表同一长度（长链表指针先遍历长度差的节点），最后比较两段链表是否相等
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode curA = headA;ListNode curB = headB;int lenA = 0, lenB = 0;while (curA != null){curA = curA.next;lenA ++;}while (curB != null){curB = curB.next;lenB ++;}curA = headA;curB = headB;if (lenB > lenA){int temp = lenA;lenA = lenB;lenB = temp;ListNode tempNode = curA;curA = curB;curB = tempNode;}int cnt = lenA - lenB;while (cnt -- > 0){curA = curA.next;}while (curA != null){if (curA == curB){return curA;}curA = curA.next;curB = curB.next;}return null;}