86. 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.

解法一：
直接插入法：

/**
 * 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) {
        if(!head || !head->next) return head;
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        ListNode* pre = dummy;
        ListNode* cur;
        while(pre->next && pre->next->val < x){
            pre = pre->next;
        }
        ListNode* head2 = pre;
        pre = head2->next;     
        while(pre&&pre->next){
            cur = pre->next;
            if(cur->val < x){
                pre->next = cur->next;  //先删除需要插入的节点
                cur->next = head2->next;
                head2->next =cur;
                head2= head2->next;
            }
            else pre = pre->next;
        }
        return dummy->next;
    }
};

解法二：
把原链表拆分成两部分，再合起来即可

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;
}