@TOC
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1、尾插法创立链表
Node head = null;public void add(int data){ Node newNode = new Node(data); if(head == null){//头结点是否为空 head = newNode; return; } Node temp = head;//查找增加位 while(temp.next != null){ temp = temp.next; } temp.next = newNode;}2、删除链表指定节点
public Boolean delete(int index){ if(index < 0 || index>length()) { return false; } //删除链表的第一个元素 if(index == 0){ head = head.next; return true; } //删除第一个以外的元素 Node pre = head; Node cur = head.next; int i=1; while(cur != null){ if(i == index){ pre.next = cur.next; return true; } pre=cur; cur=cur.next; i++;//别忘了 } return true;}3、排序链表
public Node orderList(Node head){ Node seq = null; Node cur = head; while(cur.next != null){//遍历抉择 seq = cur.next; while(seq != null) {//选出最小点 if(cur.data>seq.data){ int temp = cur.data; cur.data=seq.data; seq.data=temp; } seq=seq.next;//不要忘了，挪动到下一点持续 } cur = cur.next; } return head;}4、一般链表去重
public ListNode deleteDuplication(ListNode pHead){ if(head == null || head.next == null) { return head; } Set<Integer> set = new HashSet<Integer>(); ListNode cur = pHead; ListNode pre = null; while(cur != null) { if (!set.contains(cur.val)) { set.add(cur.val); pre = cur;//pre总指向以后增加的节点，永远指向尾部 } if(set.contains(cur.val)) { pre.next = cur.next; } cur = cur.next; } return pHead;}// 办法二public void deleteDuplecate2(){ Node cur = head; while(cur != null){ Node pre = cur; Node seq = cur.next;//这样是为了有前驱 while(seq != null){//若没有后继则不用思考去重 if(seq.data == cur.data){ pre.next = seq.next; } if(seq.data != cur.data){ pre = seq; } seq = seq.next; } cur=cur.next; }}5、排序链表去重
public void duplication(LinkList linkList) { LinkListNode cur = linkList.getHead(); LinkListNode seq = null; LinkListNode pre = null; while (cur != null) {//遍历 seq = cur.getNext(); if (seq == null) { break; } if (seq.getVal() != cur.getVal()) { cur = cur.getNext(); continue; } do { pre = seq; seq = seq.getNext(); } while (seq != null && seq.getVal() == cur.getVal()); if (seq == null) { cur.setNext(null); break; } pre.setNext(null); cur.setNext(seq); cur = seq; }}6、不晓得头指针状况下删除节点
public boolean deleteNode(Node node) { if(node==null || node.next==null) { return false; } //与后继节点替换元素 int temp = node.data; node.data = node.next.data; node.next.data = temp; node.next=node.next.next; return true;}7、宰割链表
编写代码，以给定值x为基准将链表宰割成两局部，所有小于x的结点排在大于或等于x的结点之前
public ListNode partition(ListNode pHead, int x) { // write code here if (pHead == null) { return pHead; } ListNode cur = pHead; ListNode head1 = null; ListNode pre1 = null; ListNode head2 = null; ListNode pre2 = null; while (cur != null) { if (cur.val < x && head1 == null) { head1 = cur; pre1 = cur; } if (cur.val >= x && head2 == null) { head2 = cur; pre2 = cur; } if (cur.val < x&& head1 != null) { pre1.next = cur; pre1 = cur; } if (cur.val >= x&& head1 != null) { pre2.next = cur; pre2 = cur; } cur = cur.next; } if (head1 == null) { return head2; } if (head2 == null) { return head1; } pre2.next = null;//next不是rear pre1.next = head2;//next啊亲啊 return head1;}8、快慢指针求链表环的入口点
等量关系：a+(n+1)b+nc=2(a+b) ⟹ a=c+(n−1)(b+c) ⟹ a= c + (n-1)*R式，从相遇点到入环点的间隔加上 n-1 圈的环长 R ，恰好等于从链表头部到入环点的间隔。
public ListNode detectCycle(ListNode head) { if (head == null || head.next == null) { return head; } ListNode fast = head; ListNode slow = head; boolean hasCycle = false; while(fast != null && fast.next != null) { fast = fast.next.next; slow = slow.next; if (fast == slow) { hasCycle = true; break; } } if (!hasCycle) { return null; } slow = head; while (fast != slow) { fast = fast.next; slow = slow.next; } return slow;}9、快慢指针求链表的两头节点
public ListNode searchMid(ListNode head) { ListNode slow = head, quick = head; while (quick != null && quick.next != null) { slow = slow.next; quick = quick.next.next; } return slow;}10、快慢指针判断回文链表
// 最简略的方法，通过堆栈来实现，放入栈在遍历一次
public boolean chkPalindrome(ListNode head) { // write code here if (head == null) { return false; } ListNode rear = searchMid(head); ListNode cur = rear.next ListNode seq = null; while (cur != null) { seq = cur.next;//保留后继 cur.next = rear;//逆置 rear = cur;//挪动 cur = seq;//该算法不须要返回最初节点 } while (head != rear) { if (head.val != rear.val) { return false; } if (head.next == rear && head.val == rear.val) { return true; } rear = rear.next; head = head.next; } return true;}11、求两个链表的交点
public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) { if (pHead1 == null || pHead2 == null) { return null; } ListNode tail1 = pHead1; ListNode tail2 = pHead2; while (tail1.next != null) { tail1 = tail1.next; } while (tail2.next != null) { tail2 = tail2.next; } if (tail1 != tail2) { return null; } ListNode cur1 = pHead1; ListNode cur2 = pHead2; int l1 = getLen(pHead1); int l2 = getLen(pHead2); int delta = l1 - l2; while (delta > 0) { cur1 = cur1.next; delta--; } while (delta < 0) { cur2 = cur2.next; delta++; } while (cur1 != cur2) { cur1 = cur1.next; cur2 = cur2.next; } return cur1;}12、Offer-06 从尾到头打印链表
问题形容：输出一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。输出：head = [1,3,2]，输入：[2,3,1]。
public int[] reversePrint(ListNode head) { if (head == null) { return new int[]{}; } Stack<Integer> stack = new Stack<>(); ListNode p = head; while (p != null) { stack.push(p.val); p = p.next; } int[] result = new int[stack.size()]; int k = 0; while (!stack.isEmpty()) { result[k++] = stack.pop(); } return result;}13、Offer-24 反转链表
剑指 Offer 24. 反转链表：定义一个函数，输出一个链表的头节点，反转该链表并输入反转后链表的头节点。示例:输出: 1->2->3->4->5->NULL，输入: 5->4->3->2->1->NULL
public ListNode ReverseList(ListNode head) { if (head == null || head.next == null) { return head; } ListNode pre = head, cur = head.next; ListNode seq = null, newHead = null; head.next = null; while (cur != null) { seq = cur.next; cur.next = pre; pre = cur; if (seq == null) { newHead = cur; } cur = seq; } return newHead;}14、Offer-22 链表中倒数第k个节点
剑指 Offer 22. 链表中倒数第k个节点：输出一个链表，输入该链表中倒数第k个节点。为了合乎大多数人的习惯，本题从1开始计数，即链表的尾节点是倒数第1个节点。
例如，一个链表有 6 个节点，从头节点开始，它们的值顺次是 1、2、3、4、5、6。这个链表的倒数第 3 个节点是值为 4 的节点。
public ListNode FindKthToTail(ListNode head, int k) { if (head == null || getLen(head) < k || k <= 0) { return null; } ListNode cur1 = head; ListNode cur2 = head; for (int i = 1; i < k; i++) { cur2 = cur2.next; } while (cur2.next != null) { cur2 = cur2.next; cur1 = cur1.next; } return cur1;}15、Offer-18. 删除链表的节点
剑指 Offer 18. 删除链表的节点：给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。返回删除后的链表的头节点。
示例:
输出: head = [4,5,1,9], val = 5
输入: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.
/** * 思考头结点为删除节点+非头结点删除 */public ListNode deleteNode(ListNode head, int val) { if (head == null) { return head; } if (head.val == val) { return head.next; } ListNode pre = head, cur = head.next; while (cur != null && cur.val != val) { pre = cur; cur = cur.next; } if (cur != null) { pre.next = cur.next; } return head;}16、Offer-25 合并两个排序的链表
剑指 Offer 25. 合并两个排序的链表：输出两个递增排序的链表，合并这两个链表并使新链表中的节点依然是递增排序的。示例1：输出：1->2->4, 1->3->4，输入：1->1->2->3->4->4。
public ListNode Merge(ListNode list1, ListNode list2) { if (list1 == null) { return list2; } if (list2 == null) { return list1; } ListNode head = null; if (list1.val < list2.val) { head = list1; head.next = Merge(list1.next, list2); } else { head = list2; head.next = Merge(list1, list2.next); } return head;}public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if (l1 == null) { return l2; } if (l2 == null) { return l1; } ListNode dum = new ListNode(-1); ListNode cur = dum; while (l1 != null && l2 != null) { int l1v = l1.val, l2v = l2.val; if (l1v <= l2v) { cur.next = l1; l1 = l1.next; } if (l1v > l2v) { cur.next = l2; l2 = l2.next; } cur = cur.next; } cur.next = l1 == null ? l2 : l1; return dum.next;}17、Offer-35 简单链表的复制
剑指 Offer 35. 简单链表的复制：请实现 copyRandomList 函数，复制一个简单链表。在简单链表中，每个节点除了有一个 next 指针指向下一个节点，还有一个 random 指针指向链表中的任意节点或者 null。
// 一般链表复制public Node copyRandomList(Node head) { Node cur = head; Node dum = new Node(0), pre = dum; while(cur != null) { Node node = new Node(cur.val); // 复制节点 cur pre.next = node; // 新链表的 前驱节点 -> 以后节点 // pre.random = "???"; // 新链表的 「 前驱节点 -> 以后节点 」 无奈确定 cur = cur.next; // 遍历下一节点 pre = node; // 保留以后新节点 } return dum.next;}// 简单链表复制：哈希表法public Node copyRandomList(Node head) { if (head == null) { return head; } // 复制各节点，并建设 “原节点 -> 新节点” 的 Map 映射 Node cur = head; Map<Node, Node> nodeMap = new HashMap<>(); while(cur != null) { nodeMap.put(cur, new Node(cur.val)); cur = cur.next; } cur = head; while (cur != null) { nodeMap.get(cur).next = nodeMap.get(cur.next); nodeMap.get(cur).random = nodeMap.get(cur.random); cur = cur.next; } return nodeMap.get(head); // 返回新链表的头节点}// 简单链表复制：链表合并拆分法 原节点 1 -> 新节点 1 -> 原节点 2 -> 新节点 2 ->public Node copyRandomList(Node head) { if(head == null) return null; Node cur = head; // 1. 复制各节点，并构建拼接链表 while(cur != null) { Node tmp = new Node(cur.val); tmp.next = cur.next; cur.next = tmp; cur = tmp.next; } // 2. 构建各新节点的 random 指向 cur = head; while(cur != null) { if(cur.random != null) cur.next.random = cur.random.next; cur = cur.next.next; } // 3. 拆分两链表 cur = head.next; Node pre = head, res = head.next; while(cur.next != null) { pre.next = pre.next.next; cur.next = cur.next.next; pre = pre.next; cur = cur.next; } pre.next = null; // 独自解决原链表尾节点 return res; // 返回新链表头节点}本文由mdnice多平台公布