【Partition List】cpp

题目：

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 and x = 3,
return 1->2->2->4->3->5.

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode dummyLess(-1), dummyGreaterOrEqual(-1);
        ListNode *p1 = &dummyLess, *p2 = &dummyGreaterOrEqual, *p = head;
        while(p){
            if ( p->val < x){
                p1->next = p;
                p1 = p1->next;
            }
            else{
                p2->next = p;
                p2 = p2->next;
            }
            p = p->next;
        }
        p1->next = dummyGreaterOrEqual.next;
        p2->next = NULL;
        return dummyLess.next;
    }
};

Tips:

链表的基础操作。

设立两个虚表头（代表两个子链表），分别往后接小于x的和大于等于x的，然后再把两个子链表接起来。

==================================================

第二次过这道题，稍微想了一下，想到了两个虚表头的思路；代码一次AC。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
            ListNode h1(0);
            ListNode h2(0);
            ListNode* p1 = &h1;
            ListNode* p2 = &h2;
            ListNode* p = head;
            while (p)
            {
                if ( p->val<x )
                {
                    p1->next = new ListNode(p->val);
                    p1 = p1->next;
                }
                else
                {
                    p2->next = new ListNode(p->val);
                    p2 = p2->next;
                }
                p = p->next;
            }
            p1->next = h2.next;
            return h1.next;
    }
};