0086. Partition List (M)
Partition List (M)
题目
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.
Example:
Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5
题意
给定一个链表和目标值x,将链表中所有值小于x的结点移到所有值大于等于x的结点的前面,同时不能改变值小于/大于等于x的结点的相对位置。
思路
新建两个头结点分别用来保存原链表中值小于x的结点和值大于等于x的结点,最后将两链表相连即为所求的新链表。
代码实现
Java
class Solution {
public ListNode partition(ListNode head, int x) {
// 建空的头结点便于处理
ListNode leftHead = new ListNode(0), leftTail = leftHead;
ListNode rightHead = new ListNode(0), rightTail = rightHead;
while (head != null) {
if (head.val < x) {
leftTail.next = head;
leftTail = head;
} else {
rightTail.next = head;
rightTail = head;
}
head = head.next;
}
rightTail.next = null; // 注意断链,否则可能会形成环
leftTail.next = rightHead.next; // 拼接链表
return leftHead.next;
}
}
JavaScript
/**
* @param {ListNode} head
* @param {number} x
* @return {ListNode}
*/
var partition = function (head, x) {
const dummyA = new ListNode()
const dummyB = new ListNode()
let A = dummyA
let B = dummyB
while (head) {
const next = head.next
head.next = null
if (head.val < x) {
A.next = head
A = A.next
} else {
B.next = head
B = B.next
}
head = next
}
A.next = dummyB.next
return dummyA.next
}