LeetCode_Partition List

一.题目

Partition List

   Total Accepted: 34348  Total Submissions: 124815 My Submissions

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.

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二.解题技巧

    这道题是一个普通的链表遍历的题目，需要注意的地方在于必须将链表的最后一个节点的下一个节点更新为null，不然链表会出现环，从而导致死循环的情况。

三.实现代码

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/


#include <iostream>

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};


class Solution
{
public:
    ListNode *partition(ListNode *head, int x)
    {
        ListNode *LessHeader = new ListNode(10);
        ListNode *LessTail = LessHeader;

        ListNode *GreaterHeader = new ListNode(10);
        ListNode *GreaterTail = GreaterHeader;

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

            head = head->next;
        }

        LessTail->next = GreaterHeader->next;
        GreaterTail->next = NULL;
        delete GreaterHeader;

        head = LessHeader->next;
        delete LessHeader;

        return head;
    }
};


void Test()
{
    ListNode *head = new ListNode(2);
    head->next = new ListNode(1);
    int x = 2;

    Solution S1;
    ListNode *Result = S1.partition(head, x);

    while (Result != NULL)
    {
        std::cout << "The result is " << Result->val << std::endl;
        Result = Result->next;
    }



}

四.体会

    这只是一道简单的链表遍历题，只不过在生成新的链表的时候，需要注意不要出现环，也就是记得更新最后一个元素的next指针，将其置为null，这也是这道题的一个考点。

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