[LeetCode] 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 and x = 3,
return 1->2->2->4->3->5.

思路：新建两个头结点，分别把小于x 的节点链接到第一个头结点后，把大于或等于x 的节点链接到第二个头节点后。

　　　　时间复杂度O(n),空间复杂度O(1)

/**
 * 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) {
        if(head == NULL) return head;
        
        ListNode *plow = new ListNode(-1);
        ListNode *plhead = plow;
        ListNode *phigh = new ListNode(-1);
        ListNode *phhead = phigh;
        ListNode *pt = head;
        
        while (pt != NULL) {
            if (pt->val < x) {
                plow->next = pt;
                plow = pt;
            } else {
                phigh->next = pt;
                phigh = pt;
            }
            pt = pt->next;
        }
        
        plow->next = phhead->next;
        phigh->next = NULL;
        
        return plhead->next;
    }
};