LeetCode 86. Partition List(链表题目)

LeetCode 86. Partition List(链表题目)

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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

思路分析

这里可以采用两个链表进行单独存放,只需要建立两个链表的虚拟头结点即可,其他的在判断的过程中,通过指针指向的改变就可以直接生成两个链表,这两个链表中除了两个虚拟头结点之外,其他节点均是原链表中的结点,只是被指向了新的链表而已。既然要生成两个链表,后边还要将这两个链表进行连接,那就要时刻记录链表的首尾位置,只有这样才能方便对接。

具体代码:

/**
 * 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 *dumy_big=new ListNode(0);
        ListNode *big_end=dumy_big;
        ListNode *dumy_sml=new ListNode(0);
        ListNode *sml_end=dumy_sml;
        
        while(head!=NULL)
        {
            if(head->valnext=head;
                sml_end=head;               
            }
            else
            {
                big_end->next=head;
                big_end=head;
            }         
            head=head->next;
        }
        
        if(dumy_sml->next!=NULL)
        {
            sml_end->next=dumy_big->next;
            big_end->next=NULL;
            return dumy_sml->next;
        }
        
        else
        {
            return  dumy_big->next;
        }
  
    }
};

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