@TOC
微信公众号：JavaTomStudio

1、尾插法创建链表

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;//查找添加位 
   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){//遍历选择
        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，移动到下一点继续
        }
        cur = cur.next;
    }
    return head;
}

4、普通链表去重

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总指向当前添加的节点，永远指向尾部
        }
        
        if(set.contains(cur.val)) {
            pre.next = cur.next;
        }
        cur = cur.next;
    }
    return pHead;
}
// 方法二
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){//若没有后继则不必考虑去重
            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、不知道头指针情况下删除节点

public boolean deleteNode(Node node) {
    if(node==null || node.next==null) {
        return false;
    }    
    //与后继节点交换元素
    int temp = node.data;
    node.data = node.next.data;
    node.next.data = temp;
    node.next=node.next.next;
    return true;
}

7、分割链表

编写代码，以给定值x为基准将链表分割成两部分，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距离加上 n-1 圈的环长 R ，恰好等于从链表头部到入环点的距离。


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、快慢指针求链表的中间节点

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

// 最简单的办法，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;//保存后继
        cur.next = rear;//逆置
        rear = cur;//移动
        cur = seq;//该算法不需要返回最后节点
    }
    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

问题描述：输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。输入：head = [1,3,2]，输出：[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，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。示例:输入: 1->2->3->4->5->NULL，输出: 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：输入一个链表，输出该链表中倒数第k个节点。为了符合大多数人的习惯，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，它们的值依次是 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:

输入: head = [4,5,1,9], val = 5
输出: [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.

/**
 * 考虑头结点为删除节点+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：输入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仍然是递增排序的。示例1：输入：1->2->4, 1->3->4，输出：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 复杂链表的复制

剑指 Offer 35. 复杂链表的复制：请实现 copyRandomList 函数，复制一个复杂链表。在复杂链表中，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。

在这里插入图片描述

// 普通链表复制
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 -> 当前节点
        // pre.random = "???";         // 新链表的 「 前驱节点 -> 当前节点 」 无法确定
        cur = cur.next;                // 遍历下一节点
        pre = node;                    // 保存当前新节点
    }
    return dum.next;
}

// 复杂链表复制：哈希表法
public Node copyRandomList(Node head) {
    if (head == null) {
        return head;
    }

    // 复制各节点，并建立 “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
}

// 复杂链表复制：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; // 单独处理原链表尾节点
    return res;      // 返回新链表头节点
}

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