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

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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。

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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
}