春节刷题day9：LeetCode
237. 删除链表中的节点
1290. 二进制链表转整数
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
876. 链表的中间结点
面试题 02.07. 链表相交
1、237. 删除链表中的节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode* node) {
node -> val = node -> next -> val;
node -> next = node -> next -> next;
}
};
2、1290. 二进制链表转整数
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
int getDecimalValue(ListNode* head) {
ListNode* HEAD = new ListNode(1);
while(head){
ListNode* r = head -> next;
head -> next = HEAD -> next;
HEAD -> next = head;
head = r;
}
HEAD = HEAD -> next;
int ans = 0, k = 0;
while(HEAD){
ans += (HEAD -> val) * (1 << k++);
HEAD = HEAD -> next;
}
return ans;
}
};
3、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(r -> next && cnt != k){
r = r -> next; cnt++; }
while(r -> next){
r = r -> next; l = l -> next; }
return l;
}
};
4、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* HEAD = new ListNode(1);
while(head){
ListNode* r = head -> next;
head -> next = HEAD -> next;
HEAD -> next = head;
head = r;
}
HEAD = HEAD -> next;
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;
stack<int> s;
while(head){
s.push(head -> val);
head = head -> next;
}
while(!s.empty()){
ans.push_back(s.top());
s.pop();
}
return ans;
}
};
5、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) {
ListNode* HEAD = new ListNode(1);
while(head){
ListNode* r = head -> next;
head -> next = HEAD -> next;
HEAD -> next = head;
head = r;
}
return HEAD -> next;
}
};
6、876. 链表的中间结点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* middleNode(ListNode* head) {
ListNode* ans;
ListNode* HEAD = new ListNode(1);
HEAD -> next = head;
ListNode* l = HEAD;
ListNode* r = HEAD;
while(r){
r = r -> next;
if(r -> next == NULL){
l = l -> next;
ans = l; break;
}else{
r = r -> next; l = l -> next;
}
if(r -> next == NULL){
l = l -> next; ans = l; break;
}
}
return ans;
}
};
7、面试题 02.07. 链表相交
/**
* 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;
}
};
2021/2/14完结（今天还在刷题的小伙伴一定很孤独吧，但是越努力越幸运。祝各位小伙伴情人节快乐）。