86. Partition List
86. Partition List
题目
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
题目大意
给定一个数 x,大于或等于 x 的数字都要排列在比 x 小的数字后面,并且相对位置不能发生变化。由于相对位置不能发生变化,所以不能用类似冒泡排序的思想。
解题思路
这道题最简单的做法是构造双向链表,不过时间复杂度是 O(n^2)。
更优的方法是新构造 2 个链表,一个链表专门存储比 x 小的结点,另一个专门存储比 x 大的结点。在原链表头部开始扫描一边,依次把这两类点归类到 2 个新建链表中,有点入栈的意思。由于是从头开始扫描的原链表,所以原链表中的原有顺序会依旧被保存下来。最后 2 个新链表里面会存储好各自的结果,把这两个链表,比 x 小的链表拼接到 比 x 大的链表的前面,就能得到最后的答案了。
代码
/**
* @param {ListNode} head
* @param {number} x
* @return {ListNode}
*/
var partition = function (head, x) {
let beforeHead = new ListNode(0, null);
let before = beforeHead;
let afterHead = new ListNode(0, null);
let after = afterHead;
while (head) {
if (head.val < x) {
before.next = head;
before = before.next;
} else {
after.next = head;
after = after.next;
}
head = head.next;
}
after.next = null;
before.next = afterHead.next;
return beforeHead.next;
};
相关题目
相关题目
- [2161. 根据给定数字划分数组](https://leetcode.com/problems/partition-array-according-to-given-pivot)