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.

C++实现代码：

#include<iostream>
#include<new>
using namespace std;

//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==NULL||head->next==NULL)
            return head;
        ListNode *pre=head;
        ListNode *p=head;
        ListNode *l1=NULL;
        ListNode *q=NULL;
        ListNode *k=NULL;
        while(p)
        {
            if(p->val<x)
            {
                pre=p;
                p=p->next;
            }
            else
            {
                q=p;
                if(p==head)
                {
                    pre=p->next;
                    head=p->next;
                }
                p=p->next;
                if(pre!=p)
                    pre->next=p;
                q->next=NULL;
                if(l1==NULL)
                {
                    l1=q;
                    k=q;
                }
                else
                {
                    k->next=q;
                    k=q;
                }
            }
        }
        //判断head以防要返回的链表为空
        if(head)
            pre->next=l1;
        else
            head=l1;
        return head;
    }
    void createList(ListNode *&head)
    {
        ListNode *p=NULL;
        int i=0;
        int arr[10]= {7,5,6,4,1,3,9,8,2,10};
        for(i=0; i<7; i++)
        {
            if(head==NULL)
            {
                head=new ListNode(arr[i]);
                if(head==NULL)
                    return;
            }
            else
            {
                p=new ListNode(arr[i]);
                p->next=head;
                head=p;
            }
        }
    }
};

int main()
{
    Solution s;
    ListNode *L=NULL;
    s.createList(L);
    ListNode *head=L;
    while(head)
    {
        cout<<head->val<<" ";
        head=head->next;
    }
    cout<<endl;
    L=s.partition(L,4);
    while(L)
    {
        cout<<L->val<<" ";
        L=L->next;
    }
}

运行结果：