Cinte's Leetcode Record
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92. 反转链表 II

Posted on Edited on In Disqus:
Symbols count in article: 1.5k Reading time ≈ 1 mins.

92. 反转链表 II

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/**
 * 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 pseudoHead;
        ListNode* pre = &pseudoHead;
        pre->next = head;
        int count = 1;
        while(head)
        {
            if(count == left)
                break;
            pre = head;
            head = head->next;
            ++count;
        }
        ListNode* pt = nullptr;
        while(head)
        {
            auto next = head->next;
            head->next = pt;
            pt = head;
            head = next;
            if(count++ == right)
                break;
        }
        pre->next->next = head;
        pre->next = pt;
        return pseudoHead.next;
    }
};

边遍历边顺便反转，事实上也是遍历一次，比关解方法一中先找左右再切断要好。

方法2

好绕，建议画图，head永远是那个点，会换了几次后换到最后

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/**
 * 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 pseudoHead;
        ListNode* pre = &pseudoHead;
        pre->next = head;
        int count = 1;
        while(count != left)
        {
            pre = head;
            head = head->next;
            ++count;
        }
        ListNode* next;
        while(count != right)
        {
            next = head->next;
            head->next = next->next;
            next->next = pre->next;
            pre->next = next;
            ++count;
        }
        return pseudoHead.next;
    }
};