【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) {
        ListNode dummy_left(-1), dummy_right(-1);
        ListNode *pl = &dummy_left, *pr = &dummy_right;
        for (ListNode * p = head; p != nullptr; p = p->next) {
            if (p->val < x) {
                pl->next = p;
                pl = pl->next;
            } else {
                pr->next = p;
                pr = pr->next;
            }
        }
        pl->next = dummy_right.next;
        pr->next = nullptr;
        return dummy_left.next;
    }
};

维护两个链表，一个保存小于x的部分，一个保存大于x的部分，最后将两个链表拼接起来。要求保持稳定性则应该用尾插法。加dummy元素可使代码变简单。