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.

/** * 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 || head->next == NULL)return head;ListNode lessx(0), greaterx(0);ListNode *pless = &lessx, *pgreater = &greaterx;while (head) {if (head->val < x) {pless->next = head;pless = head;}else {pgreater->next = head;pgreater = head;}head = head->next;}pless->next = greaterx.next;pgreater->next = NULL;return lessx.next;}};