反转链表

注意：反转结束后，从原来的链表上看，\(pre\) 指向反转这一段的末尾，\(cur\) 指向反转这一段后续的下一个节点。

206.反转链表

/**
 * 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) {
        ListNode* pre = nullptr, *curr = head;
        while (curr) {
            ListNode* nxt = curr->next;
            curr->next = pre;
            pre = curr;
            curr = nxt;
        }
        return pre;
    }
};

92.反转链表Ⅱ

/**
 * 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* reverseBetween(ListNode* head, int left, int right) {
        ListNode *dummy = new ListNode(0, head), *p0 = dummy;//left 为 1 时没有上一个节点，因此要维护一个哨兵节点
        for (int i = 0; i < left - 1; i++) p0 = p0->next;//结束时 p0 指向 left 的上一个节点

        ListNode *pre = nullptr, *cur = p0->next;
        for (int i = 0; i < right - left + 1; i++) {
            ListNode* nxt = cur->next;
            cur->next = pre;
            pre = cur;
            cur = nxt;
        }
        
        p0->next->next = cur;
        p0->next = pre;
        return dummy->next;
    }
};

25.K个一组翻转链表

/**
 * 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* reverseKGroup(ListNode* head, int k) {
        ListNode* cur = head;
        int len = 0;
        while (cur) {
            len++;//计算链表长度
            cur = cur->next;
        }
        ListNode *dummy = new ListNode(0, head), *p0 = dummy;
        ListNode *pre = nullptr, *curr = head;
        while (len >= k) {//剩余元素比 k 大就反转链表节
            len -= k;
            for (int i = 0; i < k; i++) {
                ListNode *nxt = curr->next;
                curr->next = pre;
                pre = curr;
                curr = nxt;
            }
            ListNode *nxt = p0->next;
            p0->next->next = curr;
            p0->next = pre;
            p0 = nxt;
        }
        return dummy->next;
    }
};