问题描述
Given the head of 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.
Example 1:
Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]
Example 2:
Input: head = [2,1], x = 2
Output: [1,2]
Constraints:
- The number of nodes in the list is in the range [0, 200].
- -100 <= Node.val <= 100
- -200 <= x <= 200
问题分析
题目要求保持原有相对顺序的基础上将单链表分为小于x、大于等于x的两部分
我们只需初始化2个指针，分别代表小于x、大于等于x部分的新链表的tail。然后遍历原单链表，各自append到对应的2个链表下。完成后，将小于x的链表再指向另外一部分的head，最终返回第一部分的head即可。
唯一需要注意的是，需要对第二部分的末尾Next进行nil处理，否则原有元素的指向会造成环，判定程序进行序列化时会TLE或者Runtime Error
示例代码
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func partition(head *ListNode, x int) *ListNode {
// 指向小于x和大于等于x两部分的指针tail
lt, gt := &ListNode{Val: 0}, &ListNode{Val: 0}
// 以上两部分的head
headlt, headgt := lt, gt
for {
if head == nil {
break
}
if head.Val < x {
lt.Next = head
lt = lt.Next
} else {
gt.Next = head
gt = gt.Next
}
head = head.Next
}
lt.Next = headgt.Next
// avoid cycle
gt.Next = nil
return headlt.Next
}