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.

基本思路：

1. 設置兩個臨時鏈表

2. 將小於x的節點，掛到一個鏈表上。將大於等於的節點，掛到另一個鏈表上。

3. 串接兩個鏈表。將後一個鏈表的頭部，掛以前一個鏈表的尾部。

所犯的錯誤：

初次提交時，漏寫了

p2->next = 0;

/**
 * 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 h1(0);
        ListNode h2(0);
        ListNode *p1 = &h1;
        ListNode *p2 = &h2;
        while (head) {
            if (head->val < x) {
                p1->next = head;
                p1 = p1->next;
            }
            else {
                p2->next = head;
                p2 = p2->next;
            }
            head = head->next;
        }
        p1->next = h2.next;
        p2->next = 0; // 此句漏寫，將會在單鏈表中產生環。
        
        return h1.next;
    }
};

在leetcode討論組中，有一個更簡潔的寫法。

https://leetcode.com/discuss/21032/very-concise-one-pass-solution

ListNode *partition(ListNode *head, int x) {
    ListNode node1(0), node2(0);
    ListNode *p1 = &node1, *p2 = &node2;
    while (head) {
        if (head->val < x)
            p1 = p1->next = head;
        else
            p2 = p2->next = head;
        head = head->next;
    }
    p2->next = NULL;
    p1->next = node2.next;
    return node1.next;
}