leetcode 链表 Partition List

Partition List

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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

.

题意：根据给定值x划分链表，使得比x小的值排在前面。要求保持元素原先的相对位置。

思路：

维护两个链表，一个是比x小的节点的链表l1，另一个是大于等于x的节点的链表l2

最后把l1的尾部接到l2的头部

复杂度：时间O(n)，空间O(1)

ListNode *partition(ListNode *head, int x) {
ListNode less(-1), larger_equal(-1);
ListNode *less_ptr = &less, *larger_equal_ptr = &larger_equal;
for (ListNode *cur = head; cur != NULL; cur = cur->next)
{
if(cur->val < x){
less_ptr = less_ptr->next = cur;
}else{
larger_equal_ptr = larger_equal_ptr->next = cur;
}
}
less_ptr->next = larger_equal.next;
larger_equal_ptr->next = NULL;
return less.next;
}