Partition List

题目初看上去比较绕，感觉有很多情况需要处理，这时就需要仔细分析问题的本质，

尽量将问题归一化。

首先根据题目的要求，比x小的元素的移动参考标准其实很固定，只需要参考链表中第一个

大于或等于x的元素记为first。线性扫描链表，依次将小于x的节点移动到first之前就可以了。

所以这里我们需要记录first的前缀节点，以及当前节点的前缀节点。

另外还有一种特殊情况，也就是头结点head的值大于或等于x。此时需要单独处理，直到head前插入了一个

小节点后，就转化为了前一种情况。

代码写得比较乱，以后会集中整理

/**
* 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) {
if(head == NULL)
return head;
int count = 0;
ListNode *truehead = head;
ListNode *first = NULL;
ListNode *nextToFirst;
ListNode *current;
ListNode *nextToCurrent;
ListNode *temp;
nextToCurrent = head;
current = head->next;
if(head->val >=x)
{

while(current != NULL)
{
if(current->val < x)
{
temp = current->next;
current->next = head;
nextToCurrent->next = temp;
truehead = current;
current = nextToCurrent->next;
count++;
first = head;
nextToFirst = truehead;
break;
}
else
{
nextToCurrent = current;
current = current->next;

}
}
}

while(current != NULL)
{
if(current->val >= x)
{
count++;
if(count == 1)
{
first = current;
nextToFirst = nextToCurrent;
}
nextToCurrent = current;
current = current->next;
}
else
{
if(first != NULL)
{
temp = current->next;
nextToFirst->next = current;
current->next = first;
nextToFirst = current;
current = temp;
nextToCurrent->next = current;
}
else
{
nextToCurrent = current;
current = current->next;
}
}
}

return truehead;
}
};

另外一种思路是重新生成两个链表，分别代表大于等于x和小于x的部分，然后将两个表拼接起来，

这种方法，实现起来非常简单，但是需要申请额外的存储空间

class Solution {
public:
ListNode *partition(ListNode *head, int x) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (head == NULL) return NULL;

ListNode* a = new ListNode(0);
ListNode* pSmall = a;
ListNode* b = new ListNode(0);
ListNode* pLarge = b;

while(head != NULL)
{
if (head->val < x)
{
pSmall->next = head;
pSmall = head;
head = head->next;
}
else
{
pLarge->next = head;
pLarge = head;
head = head->next;
}
}

pLarge->next = NULL;
pSmall->next = b->next;
return a->next;
}
};