地址：https://leetcode.com/problems/partition-list/

題目：

Given a linked list and a value $x$, partition it such that all nodes less than $x$

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

理解：

給了一個連結串列，一個x，要求把這個連結串列分為兩部分，前面是小於x的，後面是大於等於x的，且不能改變原來的相對順序。

實現：

lessPre：小於部分的尾指標
morePre：大於等於部分的尾指標
lessHead：小於部分的頭指標
moreHead：大於等於部分的頭指標

class Solution {
public:
	ListNode* partition(ListNode* head, int x) {
		ListNode* lessPre = nullptr;
		ListNode*
 
 morePre = nullptr;
		ListNode* lessHead = nullptr;
		ListNode* moreHead = nullptr;
		ListNode* p = head;
		while (p) {
			if (p->val < x) {
				if (!lessHead) {
					lessHead = p;
					lessPre = p;
				}
				else {
					lessPre->next = p;
					lessPre=lessPre->next;
				}
			}
			else {
				if (!moreHead) {
					moreHead = p;
					morePre = p;
				}
				else {
					morePre->next = p;
					morePre = morePre->next;
				}
			}
			p = p->next;
		}
		if (lessHead) {
			lessPre->next = moreHead;
			if(moreHead)
				morePre->next = nullptr;
			return lessHead;
		}
		else if (moreHead) {
			morePre->next = nullptr;
			return moreHead;
		}
		else
			return nullptr;
	}
};

可以考慮用帶頭結點的連結串列，能夠簡化很多操作。

class Solution {
public:
	ListNode* partition(ListNode* head, int x) {
		ListNode* lessHead = new ListNode(0);
		ListNode* moreHead = new ListNode(0);
		ListNode* lessPre = lessHead;
		ListNode* morePre = moreHead;
		ListNode* p = head;
		while (p) {
			if (p->val < x) {
					lessPre->next = p;
					lessPre=lessPre->next;
			}
			else {
				morePre->next = p;
				morePre = morePre->next;	
			}
			p = p->next;
		}
		lessPre->next = moreHead->next;
		morePre->next = nullptr;
		return lessHead->next;
	}
};