Partition List

 

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Total Submissions: 73252

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：依据给定值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，还有一个是大于等于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;}