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){            return head;        }        ListNode leftDummy(0);        leftDummy.next = NULL;        ListNode rightDummy(0);        rightDummy.next = NULL;        ListNode * curLeft = &leftDummy;        ListNode * curRight = &rightDummy;        ListNode * cur = head;        while(cur){            if(cur->val < x){                curLeft->next = cur;                curLeft = curLeft->next;            }else{                curRight->next = cur;                curRight = curRight->next;            }            cur = cur->next;        }        curLeft->next = NULL;        curRight->next = NULL;        if(!leftDummy.next){            return rightDummy.next;        }else{            curLeft->next = rightDummy.next;            return leftDummy.next;        }    }};