struct node{
int val;
struct node* next;
node(int v):val,next(nullptr){}
};
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
// 删除头结点
while(head && head->val == val){
ListNode* tmp = head;
head = head->next;
delete tmp;
}
// 删除非头结点
ListNode* cur = head;
while(cur && cur->next){
if(cur->next->val == val){
ListNode* tmp = cur->next;
cur->next = tmp->next;
delete tmp;
}else{
cur = cur->next;
}
}
return head;
}
};
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
ListNode* tmpHead = new ListNode();// 创建一个临时头结点
tmpHead->next = head;
ListNode* cur = tmpHead;
while(cur->next){
if(cur->next->val == val){
ListNode* tmp = cur->next;
cur->next = tmp->next;
delete tmp;
}else{
cur = cur->next;
}
}
return tmpHead->next;
}
};
class MyLinkedList {
public:
struct node{
int val;
struct node* next;
node(int v):val(v),next(nullptr){}
};
MyLinkedList() {
m_size = 0;
m_head = new node(0);
}
int get(int index) {
if(index < 0 || index > m_size-1){
return -1;
}
node* cur = m_head->next;
while(index--){
cur = cur->next;
}// 循环结束,落到目标结点
return cur->val;
}
void addAtHead(int val) {
node* newNode = new node(val);
node* tmp = m_head->next;
m_head->next = newNode;
newNode->next = tmp;
m_size++;
}
void addAtTail(int val) {
node* newNode = new node(val);
node* cur = m_head;
while(cur->next){
cur = cur->next;
}// 循环结束落到目标的前一个结点
cur->next = newNode;
newNode->next = nullptr;
m_size++;
}
// 在索引为index的结点前添加值为val的结点
// 本链表相当于数组,索引从0开始。
// 示例: addAtIndex(1,1),即在索引为1的结点前插入值为1的结点。
// m_size为当前实际有的元素数量,比最大索引小1
// 以下情况:
// index == m_size;尾插
// 0 <= index <m_size 插入到index前
// index > m_size不进行插入
// index < 0 头插
void addAtIndex(int index, int val) {
if(index < 0){
addAtHead(val);
return ;
}
if(index > m_size){
return;
}
if(index == m_size){
addAtTail(val);
return ;
}
node* newnode = new node(val);
node* cur = m_head;
while(index--){
cur = cur->next;
}// 循环结束,落到目标的前一个结点。
node* tmpNext = cur->next;
cur->next = newnode;
newnode->next = tmpNext;
m_size++;
}
void deleteAtIndex(int index) {
if(index < 0 || index > m_size-1){
return ;
}
node* cur = m_head;
while(index--){
cur = cur->next;
}// 落到目标前一个结点
node* tmp = cur->next;
cur->next = tmp->next;
delete tmp;
m_size--;
}
private:
int m_size;
struct node* m_head;
};
// 原来他妈的题目说的第几个节点是指第几个索引!!!!——来自评论区
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* tmp = nullptr;// 保存下一个结点
ListNode* cur = head;// 快指针
ListNode* pre = nullptr;// 慢指针
while(cur){
tmp = cur->next;// 保存下一个结点
cur->next = pre;// 调转方向
pre = cur;// 更新慢指针-cur==nullptr就进不来循环了,所以最后一次pre肯定被置为原来的最后一个结点。
cur = tmp;// 更新快指针
}
return pre;
}
};
class Solution {
public:
ListNode* reverse(ListNode* cur,ListNode* pre){
if(cur == nullptr)return pre;
ListNode* temp = cur->next;
cur->next = pre;
// 递归就是做了这两个工作
// pre = cur;
// cur = temp;
return reverse(temp,cur);
}
ListNode* reverseList(ListNode* head) {
***
## [24. 两两交换链表中的节点](https://leetcode.cn/problems/swap-nodes-in-pairs/)
return reverse(head,nullptr);
}
};
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* tmpHead = new ListNode(0);
tmpHead->next = head;
ListNode* cur = tmpHead;
while(cur->next && cur->next->next){
ListNode* tmp = cur->next;// 保存第一个结点
ListNode* tmp1 = cur->next->next->next;// 保存第二个结点的后面一个
cur->next = tmp->next;// 第二个变第一个
cur->next->next = tmp;// 第一个变第二个
tmp->next = tmp1;// 第二个链上之前第二个的后面一个
cur = cur->next->next;// 移动两下
// 即每次都有“哨兵”结点,第一次为人为构造的,之后为上一次遍历的第二个结点。
}
return tmpHead->next;
}
};
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* tmpHead = new ListNode(0);
tmpHead->next = head;
ListNode* fast = tmpHead;
ListNode* slow = tmpHead;
n++;// 安全起见,先加加,防止 之后再多移动一步移动空指针。
// 但实际上因为题目中给了1<=n<=sz,所以那样写也不影响。
while(n-- && fast)fast = fast->next;
while(fast){
slow = slow->next;
fast = fast->next;
}
ListNode* delNode = slow->next;
slow->next = delNode->next;
delete delNode;
return tmpHead->next;// 注意返回的是虚拟头结点的next,因为使用了虚拟头结点来统一管理所有结点。
// 刚才想了一下为什么不能返回head,因为有可能head被删掉啦。
}
};
即公共尾部
。// 时间复杂度: O(n+m)
// 空间复杂度: O(1)
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode* curA = headA;
ListNode* curB = headB;
int lenA = 0;
int lenB = 0;
while(curA){
curA = curA->next;
lenA++;
}
while(curB){
curB = curB->next;
lenB++;
}
curA = headA;
curB = headB;
if(lenB > lenA ){// 使得A为长的字符串
swap(lenA,lenB);
swap(curA,curB);
}
int gep = lenA - lenB;
while(gep--){
curA = curA->next;
}// 末尾对齐
while(curA){
if(curA == curB)return curA;
curA = curA->next;
curB = curB->next;
}
return NULL;
}
};
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(headA ==NULL || headB == NULL)return NULL;
ListNode* curA = headA;
ListNode* curB = headB;
while(curA != curB){
curA = curA == NULL ?headB:curA->next;
curB = curB == NULL ?headA:curB->next;
}
return curA;
}
};
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode* slow = head;
ListNode* fast = head;
while(fast && fast->next){
fast = fast->next->next;
slow = slow->next;
if(fast == slow){
ListNaode* index1 = fast;
ListNode* index2 = head;
while(index1 != index2){
index1 = index1->next;
index2 = index2->next;
}
return index1;
}
}
return NULL;
}
};