[Linked List]Reverse Linked List,Reverse Linked List II

一.Reverse Linked List 

(M) Reverse Linked List II (M) Binary Tree Upside Down (E) Palindrome Linked List

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *p=head,*newHead=NULL,*next=NULL;
        while(p){
            next = p->next;
            p->next = newHead;
            newHead = p;
            p = next;
        }
        return newHead;
    }
};

Next challenges: (M) Reverse Linked List II (M) Binary Tree Upside Down

 

Reverse Linked List II

Total Accepted: 58179 Total Submissions: 218404 Difficulty: Medium

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

 (E) Reverse Linked List

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode *head)
    {
        ListNode* p = head,*newHead = NULL,*next = NULL;
        while(p){
            next = p->next;
            p->next = newHead;
            newHead = p;
            p = next;
        }
        return newHead;
    }
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if(m==n){
            return head;
        }
        ListNode *m_pre=NULL,*n_next=NULL,*tail=NULL,*cur=head;
        int cnt = ;
        while(cur){//找m的前驱
            ++cnt;
            if(cnt == m){
                tail = cur;//反转之前的头，反转之后的尾
                break;
            }
            m_pre = cur;
            cur=cur->next;
        }
        while(cur){//找n的后继
            n_next = cur->next;
            if(cnt == n){
                break;
            }
            cur = n_next;
            ++cnt;
        }
        if(cur){
            cur->next = NULL;
        }
        ListNode *newHead = reverseList(tail);
        if(m_pre){
            m_pre->next = newHead;
        }else{
            head = newHead==NULL? head:newHead;
        }
        if(tail){
            tail->next = n_next;
        }
        return head;
    }
};

Next challenges: (E) Remove Duplicates from Sorted List (M) Convert Sorted List to Binary Search Tree (M) Reorder List