Partition List

Description:
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.

分析：
题意：给定一个单链表和一个x，把链表中小于x的放到前面，大于等于x的放到后面，每部分元素的原始相对位置不变。

思路：遍历一遍链表，把小于x的都挂到head1后，把大于等于x的都放到head2后，最后再把大于等于的链表挂到小于链表的后面就可以了。

#include <iostream>using namespace std;class LNode{public:    int val;    LNode *next;    LNode(int x):val(x),next(nullptr){ }};class Solution{public:    LNode* partitionList(LNode* L,int x)    {        LNode left_head(-1) ;     //头节点        LNode right_head(-1) ;    //头节点        auto left = &left_head;        auto right = &right_head;        LNode *cur = L->next;   //用来遍历链表的        while (cur != nullptr)        {            if (cur->val < x)            {                left->next = cur;                left = cur;            }            else            {                right->next = cur;                right = cur;            }            cur = cur->next;        }        left->next = right_head.next;   //连接起来        right->next = nullptr;        return left_head.next;    }};int main(){    LNode *L = new LNode(-1);   //创建头节点    LNode *p = L;    int value;    //尾插法创建一个链表,输入1->4->3->2->5->2    for (int i = 0; i < 6 ; ++i)    {        cin>>value;        p->next = new LNode(value);        p = p->next;        cout<<p->val;        if (i < 5)            cout<<"->";    }    cout<<endl;    int x = 3;    Solution solution;    LNode *q = solution.partitionList(L,x);    LNode *ptr;    //输出链表    while (q != nullptr)    {        ptr = q;        cout<<q->val;        if (q->next != nullptr)            cout<<"->";        q = q->next;        delete ptr;//释放分配的内存    }    cout<<endl;    return 0;}