86 Partition List

86 Partition List

链接：https://leetcode.com/problems/partition-list/
问题描述：
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.

Hide Tags Linked List Two Pointers

这个问题是把链表中值小于x的节点移动到前面。这里需要注意的是首先需要找到一个中间节点，需要移动的节点要移到这个中间节点的前面。可以先找到第一个值大于等于x的节点，这个节点就是中间节点。假设中间节点的前一个节点为*p，需要移动的节点是*node，

node->next=p->next;
p->next=node;
p=node;

这样就完成了移动。

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        if(head==NULL||head->next==NULL) return head;

        ListNode *t=new ListNode(INT_MIN);
        t->next=head;
        ListNode *pre=t,*p=t;

        while(p->next&&p->next->val<x)
            p=p->next;

        if(p->next==NULL) {delete t;return head;}
        head=p->next;

        while(head!=NULL)
        {
            if(head->val<x)
            {
                pre->next=head->next;
                head->next=p->next;
                p->next=head;
                p=head;
                head=pre->next;
            }
            else
            {
              pre=head;
              head=head->next;
            }

        }
        head=t->next;
        delete t;
        return head;
    }
};