LeetCode 86. Partition List(鏈表題目)
題目描述:
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
這裏可以採用兩個鏈表進行單獨存放,只需要建立兩個鏈表的虛擬頭結點即可,其他的在判斷的過程中,通過指針指向的改變就可以直接生成兩個鏈表,這兩個鏈表中除了兩個虛擬頭結點之外,其他節點均是原鏈表中的結點,只是被指向了新的鏈表而已。既然要生成兩個鏈表,後邊還要將這兩個鏈表進行連接,那就要時刻記錄鏈表的首尾位置,只有這樣才能方便對接。
具體代碼:
/**
* 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->val<x)
{
sml_end->next=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;
}
}
};