idea： 烦死了，这个题一直因为创立的指针为空，或者接入结点方法不对，结果将两个小链表搞混乱了，不过具体思路ok。将小值结点成一组，大值结点成一组，最后在首尾相连，实现起来也比较简单
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* small = new ListNode(0);
ListNode* smallHead = small;
ListNode* large = new ListNode(0);
ListNode* largeHead = large;
while (head != nullptr) {
if (head->val < x) {
small->next = head;
small = small->next;
} else {
large->next = head;
large = large->next;
}
head = head->next;
}
large->next = nullptr;
small->next = largeHead->next;
return smallHead->next;
}
};
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