LeetCode 86. Partition List 划分链表 C++

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

 有点用到了倒置链表II的方法，将符合要求的结点放置在指向pre指针的后面。这道题的思路应该是找到第一个大于等于x值的结点，他前一个位置始终定位pre指针，放置比x小的结点。

方法一（C++）

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode* dummy=new ListNode(-1);
        dummy->next=head;
        ListNode* pre=dummy,* cur=head;
        while(pre->next&&pre->next->val<x)
            pre=pre->next;
        cur=pre;
        while(cur->next){
            if(cur->next->val<x){
                ListNode* t=cur->next;
                cur->next=t->next;
                t->next=pre->next;
                pre->next=t;
                pre=t;
            }
            else
                cur=cur->next;
        }
        return dummy->next;
    }
};