LeetCode: Partition List

[b]LeetCode: Partition List[/b]

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

.

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

算法：首先，找到第一个大于等于x的节点，记其前趋节点为pre，则在继续往后遍历，若发现比x小的值，则把该节点从链表中移出来，插入到pre节点后面。代码：

/**
* 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)   return NULL;
ListNode *p = head;
ListNode *pre = NULL;
while(p && p->val < x){
pre = p;
p = p->next;
}
if(!p)
return head;
ListNode *q = NULL;
while(p->next){
if(p->next->val < x){
q = p->next;
p->next = q->next;
if(pre){
q->next = pre->next;
pre->next = q;
pre = q;
}else{
q->next = head;
head = q;
pre = q;
}
}else{
p = p->next;
}
}
return head;
}
};