Partition List

题目：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.

1->4->3->2->5->2

1->2->2->4->3->5

思路：从前往后。

代码：

class Solution {
public:
ListNode* partition(ListNode* head, int x) {
if (head==NULL||head->next==NULL){
return head;
}

ListNode *left=new ListNode(-1),*right=new ListNode(-1);
ListNode *leftTail=left,*rightTail=right;

while (head!=NULL){
if (head->val<x){
leftTail->next=head;
head=head->next;
leftTail=leftTail->next;
leftTail->next=NULL;
}else{
rightTail->next=head;
head=head->next;
rightTail=rightTail->next;
}
}
right=right->next;
leftTail->next=right;
rightTail->next=NULL;
return left->next;
//最后没有设置链表指向空，所以出错。
}
};