leetcode链表--11、partition-list（链表分区）

题目描述

 

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,Given1->4->3->2->5->2and x = 3,

return1->2->2->4->3->5

 

解题思路：.本题目是将小于x的放在左侧，大于等于x的放在后面，且相对顺序不变

1）定义两个链表   p存小于x的  q存大于等于x的

2）存完后q->next =  NULL

3）p1->next = qhead->next;

#include <iostream>
#include <malloc.h>
using namespace std;
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
    //定义两个链表  1个存小于x的  一个存大于等于x的
    //连接两个链表
    ListNode *partition(ListNode *head, int x) {
        ListNode *phead = new ListNode(0);
        ListNode *qhead = new ListNode(0);
        ListNode *p = phead;
        ListNode *q = qhead;
        while(head != NULL)
        {
            if(head->val < x)
            {
                p->next = head;
                p = p->next;
            }
            else
            {
                q->next = head;
                q = q->next;
            }
            head = head->next;
        }
        q->next = NULL;//别忘记给q之后的链表断开，否则无限循环
        p->next = qhead->next;
        return phead->next;
    }
};
ListNode *CreateList(int n)
{
    ListNode *head;
    ListNode *p,*pre;
    int i;
    head=(ListNode *)malloc(sizeof(ListNode));
    head->next=NULL;
    pre=head;
    for(i=1;i<=n;i++)
    {
        p=(ListNode *)malloc(sizeof(ListNode));
        cin>>p->val;
        pre->next=p;
        pre=p;
    }
    p->next=NULL;

    return head->next;
}
/*-------------------------输出链表-----------------------------------*/
void PrintList(ListNode *h)
{
    ListNode *p;

    p=h;//不带空的头结点
    while(p)
    {
        cout<<p->val<<" ";
        p=p->next;
        cout<<endl;
    }
}
int main()
{
    int n1;
    int x;
    ListNode *h1;
    cout<<"输入链表1的结点数目"<<endl;
    cin>>n1;
    h1 = CreateList(n1);
    cout<<"链表1为："<<endl;
    PrintList(h1);
    cout<<"输入x"<<endl;
    cin>>x;
    Solution s;
    h1 = s.partition(h1,x);
    cout<<"分区后链表1为："<<endl;
    PrintList(h1);
    return 0;
}