[LeetCode] 86. Partition List

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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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思路：定义两个指针，一个负责链接小于x的值，另一个负责链接大于等于x的值，最后把两个链表连起来。

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class Solution {public:    ListNode* partition(ListNode* head, int x) {        ListNode* l = new ListNode(-1);        ListNode* r = new ListNode(-1);        ListNode* rh = r;        ListNode* lh = l;        ListNode* p = head;        while (p != NULL) {            if (p->val < x) {                l->next = p;                l = l->next;            } else {                r->next = p;                r = r->next;            }            p = p->next;        }        r->next = NULL;        l->next = rh->next;        return lh->next;    }};