[leetcode 86] Partition List

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本文详细阐述了如何通过链表节点值将链表分为两部分，一部分包含小于给定值x的所有节点，另一部分包含大于等于x的所有节点，同时保持原始相对顺序不变。

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 small_dummy(-1);
        ListNode big_dummy(-1);
        auto small_cur = &small_dummy;
        auto big_cur = &big_dummy;
        for (auto p = head; p; p = p->next) {
            if (p->val < x) {
                small_cur->next = p;
                small_cur = p;
            } else {
                big_cur->next = p;
                big_cur = p;
            }
        }
        small_cur->next = big_dummy.next;
        big_cur->next = NULL;
        return small_dummy.next;
    }
};