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
给定一个链表和一个值x,将所有小于x值的节点放在所有大于等于x值节点的前面。但不改变节点原来的顺序。如上面的例子所示(1-4-3-2-5-2,x=3),将小于3的节点2放在前面,1-2-2-4-3-5。
思路:设置两个头指针,一个指向所有小于x的节点,一个指向所有大于等于x的节点。
/**
* 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) {
ListNode less(0),more(0);
ListNode* less_ptr=&less, *more_ptr=&more;
while(head){
if(head->val < x){
less_ptr->next = head;
less_ptr = less_ptr->next;
}
else{
more_ptr->next = head;
more_ptr = more_ptr->next;
}
head = head->next;
}
less_ptr->next = more.next;
more_ptr->next = NULL;
return less.next;
}
};