2015年8月30日星期日

Lintcode: Reverse Linked List

Reverse a linked list.
Have you met this question in a real interview? 
Yes
Example
For linked list 1->2->3, the reversed linked list is 3->2->1
Challenge
Reverse it in-place and in one-pass
Tags Expand 
Linked List

Initial Solution:
Use three pointers and draw sketch:
Don't forget to add Null to head and reverse next and cur pointer after while loop
Time: O(n)
Space: O(1)

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/**
 * Definition of ListNode
 * 
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 * 
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        // write your code here
        
        ListNode *prev = head;
        if(!head) return NULL;
        ListNode *cur = prev->next;
        if(!cur) return head;
        head->next = NULL;
        ListNode *next = cur->next;
        while(next){
            cur->next = prev;
            prev = cur;
            cur = next;
            next = next->next;
        }
        cur->next = prev;
        return cur;
    }
};
Solution 1: Exchange adjacent nodes
Two pointers solution: Set prev node as null is a convenient way to set head->next =NULL
Time: O(n)
Space: O(1)
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/**
 * Definition of ListNode
 * 
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 * 
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        // write your code here
       ListNode* prev = NULL;
       ListNode* cur = head;
       while(cur){
           ListNode* temp = cur->next;
           cur->next = prev;
           prev = cur;
           cur = temp;
       }
       return prev;
    }
};

Solution 2: Recursion
Recursive nested layers: O(n)
Time: O(n)
space:O(1) ----without stack space

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/**
 * Definition of ListNode
 * 
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 * 
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        // write your code here
       if(head == NULL) return head;
       if(head->next == NULL) return head;
       //keep tracking head of the reversed linkedlist
       ListNode *newHead = reverse(head -> next); 
       head->next->next = head;
       head->next = NULL;
       
       return newHead;
    }
};


Lintcode second:
Three pointers solution:


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class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        // write your code here
        ListNode* prev, *cur, *next;
        prev = NULL;
        cur = head;
        if(!head) return NULL;
        next = cur->next;
        while(next){
            cur->next = prev;
            prev = cur;
            cur = next;
            next = next->next;
        }
        cur->next = prev;
        return cur;
    }
};

two pointers solution:



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class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The new head of reversed linked list.
     */
    ListNode *reverse(ListNode *head) {
        // write your code here
        ListNode *prev, *cur;
        prev = NULL;
        cur = head;
        if(!head) return head;
        while(cur){
            ListNode* next = cur->next;
            cur->next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
};

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