30 Day Challenge Day 6 | Leetcode 92. Reverse Linked List II

题解

基于 Reverse Linked List 的同样思路，稍加修改，即可。但代码不够简洁，看上去过程也有些冗余。

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* prev = nullptr;
        ListNode* curr = head;
        
        while(--m) {
            prev = curr;
            curr = curr->next;
            n--;
        }
        
        ListNode *p = prev, *q = curr;
        
        prev = curr;
        curr = curr->next;
        n--;

        while(curr && n--) {
            ListNode* temp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = temp;
        }
        
        if(p) {
            p->next = prev;
            q->next = curr;
            return head; 
        }

        q->next = curr;
        return prev;
    }
};

稍加改进：用一个虚拟节点作为新的头节点。可以使得结尾的处理更简洁。

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode *dummy = new ListNode(0);        
        dummy->next = head;

        ListNode* curr = dummy;
        
        while(--m) {
            curr = curr->next;
            n--;
        }
        
        ListNode *p = curr, *q = curr->next;
        
        ListNode* prev = curr;
        curr = curr->next;

        while(curr && n--) {
            ListNode* temp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = temp;
        }

        p->next = prev;
        q->next = curr;
        return dummy->next;
    }
};