春节刷题day2：LeetCode
剑指 Offer 22. 链表中倒数第k个节点
剑指 Offer 06. 从尾到头打印链表
剑指 Offer 24. 反转链表
剑指 Offer 25. 合并两个排序的链表
剑指 Offer 52. 两个链表的第一个公共节点
剑指 Offer 18. 删除链表的节点
1、剑指 Offer 22. 链表中倒数第k个节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* getKthFromEnd(ListNode* head, int k) {
ListNode* l = head;
ListNode* r = head;
int cnt = 1;
while(cnt != k){
r = r -> next; cnt++; }
while(r -> next){
r = r -> next; l = l -> next; }
return l;
}
};
2、剑指 Offer 06. 从尾到头打印链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
vector<int> ans;
if(head == NULL) return ans;
ListNode* p = head -> next;
head -> next = NULL;
while(p){
ListNode* r = p -> next;
p -> next = head;
head = p; p = r;
}
while(head){
ans.push_back(head -> val); head = head -> next;
}
return ans;
}
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<int> reversePrint(ListNode* head) {
vector<int> ans;
if(head == NULL) return ans;
else{
ListNode* HEAD = new ListNode(1); HEAD -> next = NULL;
ListNode* p = head;
while(p){
ListNode* r = p -> next;
p -> next = HEAD -> next;
HEAD -> next = p;
p = r;
}
HEAD = HEAD -> next;
while(HEAD){
ans.push_back(HEAD -> val); HEAD = HEAD -> next;
}
return ans;
}
}
};
3、剑指 Offer 24. 反转链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head == NULL) return head;
ListNode* p = head -> next;
head -> next = NULL;
while(p){
ListNode* r = p -> next;
p -> next = head;
head = p; p = r;
}
return head;
}
};
4、剑指 Offer 25. 合并两个排序的链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL && l2 == NULL) return NULL;
else if(l1 == NULL && l2 != NULL) return l2;
else if(l1 != NULL && l2 == NULL) return l1;
ListNode* head = new ListNode(1);
ListNode* p = head;
ListNode* p1 = l1;
ListNode* p2 = l2;
while(p1 && p2){
ListNode* r1 = p1 -> next;
ListNode* r2 = p2 -> next;
if(p1 -> val <= p2 -> val){
p -> next = p1; p1 = r1;
}else{
p -> next = p2; p2 = r2;
}
p = p -> next;
}
if(p1) p -> next = p1;
else p -> next = p2;
return head -> next;
}
};
5、剑指 Offer 52. 两个链表的第一个公共节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
unordered_map<ListNode*, int> pa;
ListNode* ans = NULL;
while(headA){
pa[headA]++;
headA = headA -> next;
}
while(headB){
if(pa[headB] == 1){
ans = headB; break;
}
headB = headB -> next;
}
return ans;
}
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode* p1 = headA;
ListNode* p2 = headB;
while(p1 != p2){
p1 = p1 == NULL ? headB : p1 -> next;
p2 = p2 == NULL ? headA : p2 -> next;
}
return p1;
}
};
6、剑指 Offer 18. 删除链表的节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteNode(ListNode* head, int val) {
if(head == NULL) return head;
if(head -> val == val) return head -> next;
ListNode* pre = head;
ListNode* p = pre -> next;
while(p){
ListNode* r = p -> next;
if(p -> val == val){
pre -> next = r; p = r;
}else{
pre = pre -> next; p = p -> next;
}
}
return head;
}
};
2021/2/7完结（刷了下LeetCode上有关链表的题目，希望可以保持下手感，但是都是Easy水平的题目，再接再厉，向着Medium和Hard进发）。