【Java数据结构】你必须要掌握的链表面试经典例题（附超详细图解和代码）
目录
2，给定一个带有头结点 head 的非空单链表，返回链表的中间结点
一，写在前面
链表几乎可以说是数据结构的重中之重，链表也是大厂面试必考的知识点，要想学好链表，最重要的就是画图解决问题，如果你认为此博客写的不错的话，求点赞，求收藏，求评论，你的三连是我进步最大的动力，废话不多说，让我们学起来吧！！！
二，链表经典例题
1，反转一个单链表
public ListNode reverseList() {
if(this.head == null) {
return null;
}
ListNode cur = this.head;
ListNode prev = null;
while (cur != null) {
ListNode curNext = cur.next;
cur.next = prev;
prev = cur;
cur = curNext;
}
return prev;
}2，给定一个带有头结点 head 的非空单链表，返回链表的中间结点
public ListNode middleNode() {
if(head == null) {
return null;
}
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null){
fast = fast.next.next;
if(fast == null) {
return slow;
}
slow = slow.next;
}
return slow;
}3，输入一个链表，输出该链表中倒数第k个结点
public ListNode findKthToTail(int k) {
if(k <= 0 || head == null) {
return null;
}
ListNode fast = head;
ListNode slow = head;
while (k-1 != 0) {
fast = fast.next;
if(fast == null) {
return null;
}
k--;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
return slow;
}4，删除链表中的多个重复值
//删除所有值为key的节点
public ListNode removeAllKey(int key){
if(this.head == null) return null;
ListNode prev = this.head;
ListNode cur = this.head.next;
while (cur != null) {
if(cur.val == key) {
prev.next = cur.next;
cur = cur.next;
}else {
prev = cur;
cur = cur.next;
}
}
//最后处理头
if(this.head.val == key) {
this.head = this.head.next;
}
return this.head;
}
5，链表的回文结构
public boolean chkPalindrome(ListNode A) {
// write code here
if(head == null) return true;
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
//slow走到了中间位置-》反转
ListNode cur = slow.next;
while(cur != null) {
ListNode curNext = cur.next;
cur.next = slow;
slow = cur;
cur = curNext;
}
//反转完成
while(head != slow) {
if(head.val != slow.val) {
return false;
}
if(head.next == slow) {
return true;
}
head = head.next;
slow = slow.next;
}
return true;
}6，合并两个链表
public static ListNode mergeTwoLists(ListNode headA, ListNode headB) {
ListNode newHead = new ListNode(-1);
ListNode tmp = newHead;
while (headA != null && headB != null) {
if(headA.val < headB.val) {
tmp.next = headA;
headA = headA.next;
tmp = tmp.next;
}else {
tmp.next = headB;
headB = headB.next;
tmp = tmp.next;
}
}
if(headA != null) {
tmp.next = headA;
}
if(headB != null) {
tmp.next = headB;
}
return newHead.next;
}7，输入两个链表，找出它们的第一个公共结点。
public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if(headA == null || headB == null) {
return null;
}
ListNode pl = headA;
ListNode ps = headB;
int lenA = 0;
int lenB = 0;
while (pl != null) {
lenA++;
pl = pl.next;
}
//pl==null
pl = headA;
while (ps != null) {
lenB++;
ps = ps.next;
}
//ps==null
ps = headB;
int len = lenA-lenB;//差值步
if(len < 0) {
pl = headB;
ps = headA;
len = lenB-lenA;
}
//1、pl永远指向了最长的链表 ps 永远指向了最短的链表 2、求到了差值len步
//pl走差值len步
while (len != 0) {
pl = pl.next;
len--;
}
//同时走 直到相遇
while (pl != ps) {
pl = pl.next;
ps = ps.next;
}
return pl;
}8，判断一个链表是否有环
public boolean hasCycle() {
if(head == null) return false;
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if(fast == slow) {
return true;
}
}
return false;
}9，求有环链表的环第一个结点
public ListNode detectCycle(ListNode head) {
if(head == null) return null;
ListNode fast = head;
ListNode slow = head;
while(fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if(fast == slow) {
break;
}
}
if(fast == null || fast.next == null) {
return null;
}
fast = head;
while (fast != slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
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