[LeetCode]86. Partition List

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.

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/** * 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 NULL;        ListNode* pCur = head;        ListNode* left = new ListNode(-1);        ListNode* right = new ListNode(-1);        ListNode* l = left;        ListNode* r = right;        while (pCur){            if (pCur->val < x){                l->next = pCur;                l = l->next;                //pCur_left = NULL;            }            else{                r->next = pCur;                r = r->next;            }            pCur = pCur->next;        }        // r的next指针必须置为0，因为它是right的最后一个结点        r->next = NULL;        l->next = right->next;        return left->next;    }};