NO CODING NO LIFE: Lintcode: Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2->null and x = 3,
return 1->2->2->4->3->5->null.

Have you met this question in a real interview?

Yes

Example

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Two Pointers Linked List

Set two dummy head nodes, put nodes into two head by value
Note: Line 40: Don't forget to set the next of last node is NULL
http://www.cnblogs.com/springfor/p/3862392.html
Time: O(n)
Space: O(1)

/**
 * 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.
     * @param x: an integer
     * @return: a ListNode 
     */
    ListNode *partition(ListNode *head, int x) {
        // write your code here
        ListNode* head1 = new ListNode(0);
        ListNode* head2 = new ListNode(0);
        ListNode* p1 = head1;
        ListNode* p2 = head2;
        
        ListNode* cur = head;
        while(cur != NULL){
            if(cur->val < x){
                p1->next = cur;
                p1 = p1->next;
                cur = cur->next;
            }
            else{
                p2->next = cur;
                p2 = p2->next;
                cur = cur->next;
            }
        }
        p2->next = NULL;
        p1->next = head2 ->next;
        return head1->next;
    }
};

Lintcode second :

/**
 * 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.
     * @param x: an integer
     * @return: a ListNode 
     */
    ListNode *partition(ListNode *head, int x) {
        // write your code here
        ListNode *p1, *p2, *p3, *p4;
        p1 = new ListNode(0);
        p2 = new ListNode(0);
        p3 = p1;
        p4 = p2;
        while(head){
            if(head->val < x) {
                p1->next = head;
                p1 = p1->next;
            }
            else {
                p2->next = head;
                p2 = p2->next;
            }
            head = head->next;   
        }
        
        p1->next = p4->next;
        head = p3->next;
        p2->next = NULL; //Set the end node points to NULL
        return head;
    }
};

NO CODING NO LIFE

2015年8月30日星期日

Lintcode: Partition List

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