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.

SOLUTION 1:

1.将链表分逐个遍历，按大小分别加入ListL和ListR；

2.将ListL和ListR连接起来；

注意：1.ListL和ListR加入dummyNode可减少判断次数；

　　　2.连接时ListL为空的情况；

　　　3.ListR的tail->next要置为NULL；

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *partition(struct ListNode *head, int x) {
    struct ListNode *headL = ListNode();
    struct ListNode *tailL = headL;

    struct ListNode *headR = ListNode();
    struct ListNode *tailR = headR;

    while(head) {
        if(head->val < x) {
            tailL->next = head;
            tailL = tailL->next;
        }
        else {
            tailR->next = head;
            tailR = tailR->next;
        }
        head = head->next;

    }
//set head must follow the join list step for the case listL is null
    tailL->next = headR->next;
    head = headL->next;
    free(headL);
    free(headR);
    tailR->next = NULL;
    return head;
}