链表
206. 反转链表
给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
示例 1：
输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]
示例 2：
输入：head = [1,2]
输出：[2,1]
示例 3：
输入：head = []
输出：[]
提示：
链表中节点的数目范围是 [0, 5000]
-5000 <= Node.val <= 5000
class Solution {
public:ListNode* reverseList(ListNode* head) {ListNode* prev = nullptr;ListNode* curr = head;while (curr) {ListNode* next = curr->next;curr->next = prev;prev = curr;curr = next;}return prev;}
};
148. 排序链表
给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。
示例 1：
输入：head = [4,2,1,3]
输出：[1,2,3,4]
示例 2：
输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]
示例 3：
输入：head = []
输出：[]
提示：
链表中节点的数目在范围 [0, 5 * 104] 内
-105 <= Node.val <= 105
class Solution {
public:ListNode* sortList(ListNode* head) {if (head == nullptr) {return head;}int length = 0;ListNode* node = head;while (node != nullptr) {length++;node = node->next;}ListNode* dummyHead = new ListNode(0, head);for (int subLength = 1; subLength < length; subLength <<= 1) {ListNode* prev = dummyHead, *curr = dummyHead->next;while (curr != nullptr) {ListNode* head1 = curr;for (int i = 1; i < subLength && curr->next != nullptr; i++)curr = curr->next;ListNode* head2 = curr->next;curr->next = nullptr;curr = head2;for (int i = 1; i < subLength && curr != nullptr && curr->next != nullptr; i++) {curr = curr->next;} ListNode* next = nullptr;if (curr != nullptr) {next = curr->next;curr->next = nullptr;}ListNode* merged = merge(head1, head2);prev->next = merged;while (prev->next != nullptr) {prev = prev->next;}curr = next;}}return dummyHead->next;}ListNode* merge(ListNode* head1, ListNode* head2) {ListNode* dummyHead = new ListNode(0);ListNode* temp = dummyHead, *temp1 = head1, *temp2 = head2;while (temp1 != nullptr && temp2 != nullptr) {if (temp1->val <= temp2->val) {temp->next = temp1;temp1 = temp1->next;} else {temp->next = temp2;temp2 = temp2->next;}temp = temp->next;}if (temp1 != nullptr) {temp->next = temp1;} else if (temp2 != nullptr) {temp->next = temp2;}return dummyHead->next;}
};
面试题 02.05. 链表求和
给定两个用链表表示的整数，每个节点包含一个数位。
这些数位是反向存放的，也就是个位排在链表首部。
编写函数对这两个整数求和，并用链表形式返回结果。
示例：
输入：(7 -> 1 -> 6) + (5 -> 9 -> 2)，即617 + 295
输出：2 -> 1 -> 9，即912
进阶：思考一下，假设这些数位是正向存放的，又该如何解决呢?
示例：
输入：(6 -> 1 -> 7) + (2 -> 9 -> 5)，即617 + 295
输出：9 -> 1 -> 2，即912
class Solution {
public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {ListNode* ans = new ListNode(0);auto now = ans;int c = 0, num;while(l1 && l2) {num = l1->val + l2->val+c;ListNode* temp = new ListNode(num%10);now->next = temp;now = now->next;c = num/10;l1 = l1->next;l2 = l2->next;}while(l1) {num = l1->val+c;ListNode* temp = new ListNode(num%10);now->next = temp;now = now->next;c = num/10;l1 = l1->next;}while(l2) {num = l2->val+c;ListNode* temp = new ListNode(num%10);now->next = temp;now = now->next;c = num/10;l2 = l2->next;}if (c) {ListNode* temp = new ListNode(c);now->next = temp;now = now->next;}return ans->next;}
};
