leetcode：Reverse Linked List

Reverse a singly linked list.

 

代码如下：

the iterative solution：（c++）

/**
 * 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 *temp = NULL , *nextNode = NULL;
        while(head){
            nextNode = head->next;   //save the current's next node 
            head->next = temp;      //let the current point to its previous one
            temp = head;           //save the current node as pre
            head = nextNode;      //move to next node
        }
        return temp;            //just point to the last node we wanted
    }
};

　或：

class Solution {
    public:
        ListNode *reverseList(ListNode *head)
{
    if (head == NULL || head->next == NULL)
        return head;

    ListNode *pCurr = head;
    ListNode *pPrev = NULL;
    ListNode *pNext = NULL;

    while (pCurr != NULL)
    {
        pNext = pCurr->next;  //save next node
        pCurr->next = pPrev;
        if (pNext == NULL)
            head = pCurr;
        pPrev = pCurr;
        pCurr = pNext;
    }

    return head;
}
    };

　　

　

其他解法：

 1、 the recursive version：（c++）

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL || head->next == NULL) return head;   // case proof and recursion exit conditon
        ListNode *np = reverseList(head->next);
        head->next->next = head;         // make the next node point back to the node itself
        head->next = NULL;               // destroy the former pointer
        return np;
    }
};

　2、（c++）

 class Solution {
    public:
        ListNode* reverseList(ListNode* head) {
            stack<ListNode*> s;
            ListNode *tail=NULL;
            while(head)
            {
                s.push(head);
                head=head->next;
            }
            if(!s.empty())
            head=s.top();
            while(!s.empty())
            {
                tail=s.top();
                s.pop();
                if(!s.empty())
                tail->next=s.top();
                else
                tail->next=NULL;
            }
            return head;
        }
    };

　　3、（c）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) 
{
        struct ListNode* last = 0;

        while (head)
        {
            struct ListNode* next = head->next;
            head->next = last;
            last = head;
            head = next;
        };

        return last;    
}

　或：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) {
    if (!head)
        return NULL;

    struct ListNode *curr = head;
    struct ListNode *next = NULL;
    struct ListNode *prev = NULL;

    while (curr){
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
        head = prev;
    }

    return head;
}

　　

　更多：http://blog.chinaunix.net/uid-7471615-id-83821.html