LeetCode 86. 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.

answer:

class Solution {public:    ListNode* partition(ListNode* head, int x) {        ListNode nodeLess(0), nodeGreater(0);        ListNode * less = & nodeLess, * greater = &nodeGreater;        while(head){            if(head->val < x){                less->next = head;                less = less->next;            }            else{                greater->next = head;                greater = greater->next;            }            head = head->next;        }        greater->next = NULL;        less->next = nodeGreater.next;        return nodeLess.next;    }};