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文章目录
- 前言
- [一、力扣21. 合并两个有序链表](#一、力扣21. 合并两个有序链表)
- [二、力扣86. 分隔链表](#二、力扣86. 分隔链表)
- [三、力扣23. 合并 K 个升序链表](#三、力扣23. 合并 K 个升序链表)
- [四、力扣19. 删除链表的倒数第 N 个结点](#四、力扣19. 删除链表的倒数第 N 个结点)
- [五、力扣876. 链表的中间结点](#五、力扣876. 链表的中间结点)
- [六、力扣142. 环形链表 II](#六、力扣142. 环形链表 II)
- [七、力扣160. 相交链表](#七、力扣160. 相交链表)
前言
一、力扣21. 合并两个有序链表
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode head = new ListNode(-1,null);
ListNode p1 = list1, p2 = list2, r = head;
while(p1 != null && p2 != null){
if(p1.val <= p2.val){
r.next = p1;
p1 = p1.next;
r = r.next;
}else{
r.next = p2;
p2 = p2.next;
r = r.next;
}
}
if(p1 != null){
r.next = p1;
}
if(p2 != null){
r.next = p2;
}
return head.next;
}
}二、力扣86. 分隔链表
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode p1 = new ListNode(), p2 = new ListNode();
ListNode r1 = p1, r2 = p2, r = head;
while(r != null){
if(r.val < x){
ListNode temp = r;
r = r.next;
temp.next = null;
r1.next = temp;
r1 = r1.next;
}else{
ListNode temp = r;
r = r.next;
temp.next = null;
r2.next = temp;
r2 = r2.next;
}
}
r1.next = p2.next;
return p1.next;
}
}三、力扣23. 合并 K 个升序链表
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode head = new ListNode(), r = head;
if(lists.length == 0){
return null;
}
PriorityQueue<ListNode> pq = new PriorityQueue<>((a,b)->{
return a.val - b.val;
});
for(ListNode node : lists){
if(node != null){
pq.add(node);
}
}
while(! pq.isEmpty()){
ListNode cur = pq.poll();
r.next = cur;
cur = cur.next;
r = r.next;
if(cur != null){
pq.add(cur);
}
}
return head.next;
}
}四、力扣19. 删除链表的倒数第 N 个结点
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode res = new ListNode(-1,head);
ListNode r1 = res, r2 = res;
int len = 0;
while(len < n+1){
r1 = r1.next;
len ++;
}
while(r1 != null){
r1 = r1.next;
r2 = r2.next;
}
r2.next = r2.next.next;
return res.next;
}
}五、力扣876. 链表的中间结点
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode middleNode(ListNode head) {
ListNode p1 = head, p2 = head;
while(p2 != null && p2.next != null){
p2 = p2.next;
p1 = p1.next;
if(p2 != null){
p2 = p2.next;
}else{
break;
}
}
return p1;
}
}六、力扣142. 环形链表 II
java
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head == null)return null;
ListNode fast = head, slow = fast;
while(fast != null){
slow = slow.next;
fast = fast.next;
if(fast == null){
return null;
}else{
fast = fast.next;
if(fast == slow){
slow = head;
while(slow != fast){
slow = slow.next;
fast = fast.next;
}
return fast;
}
}
}
return null;
}
}七、力扣160. 相交链表
java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode p1 = headA, p2 = headB;
ListNode max, min;
int lenA = 0, lenB = 0, edge = 0;
while(p1 != null){
lenA ++;
p1 = p1.next;
}
while(p2 != null){
lenB ++;
p2 = p2.next;
}
if(lenA > lenB){
edge = lenA - lenB;
max = headA;
min = headB;
}else{
edge = lenB - lenA;
max = headB;
min = headA;
}
while(edge > 0){
edge --;
max = max.next;
}
while(max != null && min != null){
if(max == min){
return max;
}
max = max.next;
min = min.next;
}
return null;
}
}