86. Partition List

1.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.

来自 https://leetcode.com/problems/partition-list/description/

2.题目分析

要求把链表中把>=x的元素放到链表末尾，逐个比较，如果符合条件则从链表中将此元素删除，并将删除的元素用尾插法加入新的链表。直到遍历完链表，再将两个链表连接起来。

3.C++代码

//我的代码：（beats 87%）
ListNode* partition(ListNode* head, int x)
{
    if (head == NULL)return head;
    ListNode*dummy = new ListNode(0);
    dummy->next = head;
    ListNode*p = dummy;
    ListNode*r = new ListNode(0);//新链表存储>=x的值
    ListNode*tail = r;
    while (p->next)
    {
        if (p->next->val >= x)
        {
            tail->next = p->next;
            p->next = p->next->next;
            tail = tail->next;
            tail->next = NULL;
        }
        else
            p = p->next;
    }
    p->next = r->next;//链表连接
    return dummy->next;
}

//另一种方法：（beats 87%）
//分两个区间left和right；
ListNode* partition2(ListNode* head, int x)
{
    if (head == NULL)return head;
    ListNode*leftdummy = new ListNode(0);
    ListNode*rightdummy = new ListNode(0);
    ListNode*leftcur = leftdummy;
    ListNode*rightcur = rightdummy;
    while (head)
    {
        if (head->val < x)
        {
            leftcur->next = head;
            leftcur = leftcur->next;
        }
        else
        {
            rightcur->next = head;
            rightcur = rightcur->next;
        }
        head = head->next;
    }
    leftcur->next = rightdummy->next;