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导师贪腐出逃美国，两年未归，可怜了我。拿了小米和美团的offer，要被延期，offer失效，工作重新找。把准备过程纪录下来，共勉。

链表是最常考察的数据结构

// 链表定义
public class Node{
    int val;//数据域
    Node next;// 指针域
    
    public Node(int val){
        this.val = val;
    }
}

1. 基础题

   • 求单链表的长度
     考察链表遍历的方法，时间复杂度为o(n)

   //求单链表长度，起手判断链表是否为空，非常必要
   public static int getListLength(Node head){
       //如果链表为空，长度为0
       if(null == head){
           return 0;
       }
       // 开始着手办正事了
       int len = 0;
       Node cur = head;
       //如果循环判断条件不确定，可以先写逻辑，回头再定条件
       while(null != cur.next){
           len++;
           cur = cur.next;
       }
       return len;
   }

2. 结点相对位置关系，结点与链表的相对位置关系

   • 翻转链表
     链表中经典解法1，引入两个指针，然后确定其相对位置关系

     • 迭代法
       解题思路：前后两个指针，遍历链表，每遍历一个结点，前面的指针将指向后面的指针，时间复杂度o(n)

   public static Node getReverseList(Node head){
       if(null == head || null == head.next){
           return null;
       }
       
       Node cur = head;
       Node newHead = null;
       
       while(null != cur){
           Node preCur = cur;
           cur = cur.next;
           preCur.next = newHead;
           newHead = preCur;
       }
       
       return newHead;
   }

   • 返回单链表倒数第k个结点
     两个指针，前面的指针超前后面指针k个位置

   public static Node findLastKthNode(Node head,int k){
       if(k == 0 || null == head){
           return null;
       }
       
       Node p = head;
       Node q = head;
       
       while(k > 0 || p != null){
           p = p.next;
           k--;
       }
       
       if(k > 0 || p == null){
           return null;
       }
       
       while(p != null){
           q = q.next;
           p = p.next;
       }
       
       return q;
   }

   • 查找中间结点
     分析：两个指针可以解决，位置相对关系是：p指针向前移动一步，q指针向前移动两步

   public static Node getMiddleNode(Node head){
       if(null == head || head.next == null){
           return head;
       }
       
       Node p = head;
       Node q = head;
       
       while(p != null){
           p = p.next;
           q = q.next;
           if(p != null){
               p = p.next;
           }
       }
       
       return q;
   }

   • 从尾到头打印单链表

   public static void printList(Node head){
       Stack<Node> s = new Stack<>();
       
       Node cur = head;
       while(cur != null){
           s.push(cur);
           cur = cur.next;
       }
       
       while(!s.empty()){
           cur = s.pop();
           System.out.print(p.val + " ");
       }
       
       System.out.println();
   }

3. 有环问题

   • 判断一个单链表是否有环
     思路：快慢两个指针，如果相遇，则有环

   public static boolean hasCycle(Node head){
       Node fast = head;
       Node slow = head;
       
       while(fast != null && fast.next != null){
           fast = fast.next.next;
           slow = slow.next;
           if(fast == slow){
               return true;
           }
       }
       
       return false;
   }

   - **取出有环链表中，环的长度**

   思路：找到相遇的结点a后，让b结点从a位置向下遍历，回来后得到的结点数，即为环长度

   //方法：有环链表中，获取环的长度。参数node代表的是相遇的那个结点  
   public int getCycleLength(Node node) { 
       if(node == null){
           return 0;
       }
       
       int length = 0;
       Node cur = node;
       while(cur != null){
           cur = cur.next;
           length++;
           if(cur != node){
               return length;
           }
       }
       
       return length;
   }

   • 判断两个链表相交
     思路：如果两链接的尾结点相同，则有环。注意，有环的链表，此种方法失效。

   public static boolean isIntersect(Node head1, Node head2) {
       Node tail1 = head1;
       Node tail2 = head2;
       
       while(tail1 != null){
           tail1 = tail1.next;
       }
       
       while(tail2 != null){
           tail2 = tail2.next;
       }
       
       if(tail1 == tail2){
           return true;
       }
       
       return false;
   }  

   • 单链表中，取出环的起始点

   public Node getCycleStart(Node head, int cycleLength) { 
       Node first = head;
       Node second = head;
       
       for(int i = 0; i < cycleLength; i++){
           second = second.next;
       }
       
       while(first != null && second != null){
           if(first == second){
               return first;
           }
           first = first.next;
           second = second.next;
       }
       
       return null;
   }

   • 求两个单链表相交的第一个节点
     思路：分别求出a链表、b链表的长度len1、len2，如果a链表比b链表长，则a提前移动len1-len2距离，然后a、b一起向前移动，直到引用相同。

   public static Node getFirstCommonNode(Node head1, Node head2) {
   
   
       Node tail1 = head1;
       Node tail2 = head2;
       
       int len1 = 1;
       while(tail1 != null){
           tail1 = tail1.next;
           len1++;
       }
       
       int len2 = 1;
       while(tail2 != null){
           tail2 = tail2.next;
           len2++;
       }
       
       if(tail1 != tail2){
           return null;
       }
       
       Node n1 = head1;
       Node n2 = head2;
       
       if(len1 > len2){
           int k = len1 - len2;
           
           while(k != 0){
               n1 = n1.next;
               k--;
           }
       }
       
       if(len1 < len2){
           int k = len1 - len2;
           
           while(k != 0){
               n2 = n2.next;
               k--;
           }
       }
       
       while(n1 != n2){
           n1 = n1.next;
           n2 = n2.next;
       }
       
       return n1;
   }  

   • 已知两个单链表head1和head2各自有序，把它们合并成一个链表依然有序

   public static Node mergeSortedList(Node head1, Node head2){
       if(head1 == null){
           return head2;
       }
       
       if(head2 == null){
           return head1;
       }
       
       Node mergeHead = null;
       if(head1.val < head2.val){
           mergeHead = head1;
           head1 = head1.next;
           mergeHead.next = null;
       }else{
           mergeHead = head2;
           head2 = head2.next;
           mergeHead.next = null;
       }
       
       Node mergeCur = mergeHead;
       while(head1 != null && head2 != null){
           if(head1.val < head2.val){
               mergeCur.next = head1;
               head1 = head1.next;
               mergeCur = mergeCur.next;
               mergeCur.next = null;
           }else{
               mergeCur.next = head2;
               head2 = head2.next;
               mergeCur = mergeCur.next;
               mergeCur.next = null;
           }
       }
       
       if(head1 != null){
           mergeCur.next = head1; 
       }
       
       if(head2 != null){
           mergeCur.next = head2;
       }
       
       return mergeHead;
   }