92. 反转链表 II EL1S

顺序：橙色->红色->蓝色->紫色

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        auto left = dummy, right = head;
        int cnt = m - 1;
        while(cnt--)
        {
            left = right;
            right = right->next;
        }
        auto savehead = left;
        auto savetail = right;
        left = right;
        right = right->next;
        cnt = n - m;
        while(cnt--) 
        {
            auto nxt = right->next;
            right->next = left;
            left = right;
            right = nxt;
        }
        savehead->next = left;
        savetail->next = right;
        return dummy->next;
    }
};

递归

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if(!head || m == n) return head;
        auto dummy = new ListNode(0);
        dummy->next = head;
        auto prev = dummy;
        int move = m - 1;
        while(move--)
        {
            prev = prev->next;
        }
        auto tail = prev->next;
        move = n - m;
        while(move--)
        {
            auto tmp = tail->next;
            tail->next = tmp->next;
            tmp->next = prev->next;
            prev->next = tmp;
        }
        return dummy->next;

    }
};

第二次写题

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* prev = dummy;
        for(int i = 0; i < m - 1; i++)
            prev = prev->next;
        ListNode *tail = prev->next;
        // cout << prev->val << endl;
        for(int i = m; i < n; i++)
        {
            ListNode* cur = tail->next;
            tail->next = cur->next;
            cur->next = prev->next;
            prev->next = cur;
        }
        return dummy->next;
    }
};