[leetcode] 86. Partition List

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.

For example,

Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

这道题是按给定值分割链表，小于给定值的在前，不小于给定值的在后，保持原来的相对顺序，题目难度为Medium。

对于任意节点，如果小于给定值则不做处理，直接遍历下一节点，如果不小于给定值则将其取出依次放置到新链表尾部，然后修改它之前节点的next指针指向它的下一个节点，最终将新的链表接在原链表尾部即可。整体比较简单，关键在于链表头的处理上。具体代码：

class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* cur = head;
ListNode* lPre = NULL;
ListNode* rPre = NULL;
ListNode* rHead = NULL;
while(cur) {
if(cur->val < x) {
lPre = cur;
}
else {
if(!lPre) head = cur->next;
else lPre->next = cur->next;
if(!rPre) rHead = cur;
else rPre->next = cur;
rPre = cur;
}
cur = cur->next;
}
if(lPre) lPre->next = rHead;
else head = rHead;
if(rPre) rPre->next = NULL;
return head;
}
};