目录
- 1 介绍
- 2 训练
- 3 参考
1 介绍
本博客用来记录代码随想录leetcode200题中链表部分的题目。
2 训练
题目1：203移除链表元素
C++代码如下，
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeElements(ListNode* head, int val) {ListNode* p = new ListNode(0, head);ListNode* res = p;while (p->next != nullptr) {if (p->next->val == val) p->next = p->next->next;else p = p->next;}return res->next;}
};
python3代码如下，
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:p = ListNode(0, head)res = pwhile p.next is not None:if p.next.val == val:p.next = p.next.next else:p = p.next return res.next
题目2：707设计链表
C++代码如下，
// struct ListNode {
// int val;
// ListNode* next;
// ListNode() : val(0), next(nullptr) {}
// ListNode(int x) : val(x), next(nullptr) {}
// ListNode(int x, ListNode* nx) : val(x), next(nx) {}
// };class MyLinkedList {
private:ListNode* head; //head无实际含义public:MyLinkedList() {head = new ListNode(0);}int get(int index) {//时间复杂度O(N)int i = 0;ListNode* p = head->next;while (p != nullptr) {if (i == index) return p->val;p = p->next;i += 1;}return -1;}void addAtHead(int val) {//时间复杂度O(1)ListNode* p = new ListNode(val, head->next);head->next = p; }void addAtTail(int val) {//时间复杂度O(N)ListNode* p = head;while (p->next != nullptr) p = p->next;ListNode* t = new ListNode(val);p->next = t;}void addAtIndex(int index, int val) {//时间复杂度O(N)ListNode* p = head; //p从头结点开始算int i = 0;while (p != nullptr && i < index) {p = p->next;i += 1;}if (p == nullptr) return;ListNode* t = new ListNode(val, p->next);p->next = t;}void deleteAtIndex(int index) {//时间复杂度为O(N)ListNode* p = head; //p从头结点开始算int i = 0;while (p != nullptr && i < index) {p = p->next;i += 1;}if (p == nullptr || p->next == nullptr) return;p->next = p->next->next;}
};/*** Your MyLinkedList object will be instantiated and called as such:* MyLinkedList* obj = new MyLinkedList();* int param_1 = obj->get(index);* obj->addAtHead(val);* obj->addAtTail(val);* obj->addAtIndex(index,val);* obj->deleteAtIndex(index);*/
python3代码如下，
class MyLinkedList:def __init__(self):self.head = ListNode(0)def get(self, index: int) -> int:#时间复杂度为O(N)i = 0p = self.head.next while p is not None:if i == index:return p.val p = p.next i += 1return -1def addAtHead(self, val: int) -> None:#时间复杂度为O(1)p = ListNode(val, self.head.next)self.head.next = pdef addAtTail(self, val: int) -> None:#时间复杂度为O(N)p = self.head while p.next is not None:p = p.next t = ListNode(val)p.next = t def addAtIndex(self, index: int, val: int) -> None:#时间复杂为O(N)p = self.head i = 0while p is not None and i < index:p = p.next i += 1if p is None:return t = ListNode(val, p.next)p.next = t def deleteAtIndex(self, index: int) -> None:#时间复杂度为O(N)p = self.head i = 0while p is not None and i < index:p = p.next i += 1if p is None or p.next is None:return p.next = p.next.next # Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)
题目3：206. 反转链表
C++代码如下，
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {if (head == nullptr || head->next == nullptr) { //特判空集或一个结点的时候return head;}ListNode* a = head;ListNode* b = a->next;while (a != nullptr && b != nullptr) {ListNode* t = b->next;b->next = a;a = b;b = t;}head->next = nullptr;return a;}
};
python3代码如下，
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:if head is None or head.next is None: #特判空集和一个结点的情况return head a = head b = a.next while a is not None and b is not None:t = b.nextb.next = a a = bb = t head.next = None return a
题目4：24两两交换链表中的节点
C++代码如下，
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* swapPairs(ListNode* head) {if (head == nullptr || head->next == nullptr) { //特判空集或者只有一个结点的情况return head;}ListNode* a = head;ListNode* b = a->next;ListNode* res = b;ListNode* aprev = nullptr;while (a != nullptr && b != nullptr) {ListNode* t = b->next;b->next = a;a->next = t;if (aprev != nullptr) aprev->next = b;aprev = a;a = t;if (a != nullptr) b = a->next;else break;}return res;}
};
python3代码如下，
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:if head is None or head.next is None: #特判空集或者一个结点的情况return head a = head b = a.next aprev = None res = b while a is not None and b is not None:t = b.next b.next = a a.next = t if aprev is not None:aprev.next = baprev = a a = tif a is not None:b = a.next else:breakreturn res
题目5：19删除链表的倒数第 N 个结点
C++代码如下，
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeNthFromEnd(ListNode* head, int k) {//删除倒数第k个结点，快慢指针法ListNode* dummy = new ListNode(0, head);ListNode* a = dummy;ListNode* b = dummy;while (k--) { //b先走k步b = b->next;}while (b->next != nullptr) { //b走到最后一个结点a = a->next;b = b->next;}if (a->next != nullptr) a->next = a->next->next;else cout << "a->next is nullptr!" << endl;return dummy->next;}
};
python3代码如下，
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def removeNthFromEnd(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:dummy = ListNode(0, head)a = dummyb = dummy while k > 0:b = b.nextk -= 1while b.next is not None:b = b.next a = a.next a.next = a.next.next return dummy.next
题目6：面试题 02.07. 链表相交
C++代码如下，
/*** 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) {if(headA == nullptr || headB == nullptr) { //特判都为空集，或者有一方为空集的情况return nullptr;}ListNode* a = headA;ListNode* b = headB;bool flaga = true;bool flagb = true;while (a != b) {a = a->next;b = b->next;if (a == nullptr) {if (flaga) {a = headB;flaga = false;} else {break;}}if (b == nullptr) {if (flagb) {b = headA;flagb = false;} else {break;}}}if (a == b) return a;else return nullptr;}
};
python3代码如下，
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = Noneclass Solution:def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:if headA is None or headB is None:return None a = headA b = headB flaga = Trueflagb = True while a != b:a = a.next b = b.next if a is None:if flaga:a = headB flaga = Falseelse:break if b is None:if flagb:b = headA flagb = False else:break if a == b:return a else:return None题目7：142. 环形链表 II
C++代码如下，
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode *detectCycle(ListNode *head) {//这次的快慢指针，是速度上的快慢指针，而非提前走的快慢指针ListNode* a = head;ListNode* b = head;while (b != nullptr && b->next != nullptr) {a = a->next;b = b->next->next;//起点到环入口的距离为x//快慢指针相遇点到环入口的距离为y//环的周长为c//有x % c = y恒成立if (a == b) { ListNode* t = head;while (t != a) {t = t->next;a = a->next;}return a;}}return nullptr;}
};
python3代码如下，
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = Noneclass Solution:def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:a = head b = head while b is not None and b.next is not None:a = a.next b = b.next.next if a == b:t = head while t != a:t = t.next a = a.next return a return None 3 参考
代码随想录官网
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