力扣挑战赛第15天-No.86分隔链表

题目描述

给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。

你应当 保留 两个分区中每个节点的初始相对位置。

示例：

输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]

输入：head = [2,1], x = 2
输出：[1,2]

注意：

链表中节点的数目在范围 [0, 200] 内
-100 <= Node.val <= 100
-200 <= x <= 200

解法一

模拟

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
let partition = function(head, x) {
  let smallHead = new ListNode();
  let largeHead = new ListNode();
  let small = smallHead;
  let large = largeHead;
  while (head) {
    if (head.val < x) {
      small.next = head;
      small = small.next;
    } else {
      large.next = head;
      large = large.next;
    }
    head = head.next;
  }
  large.next = null;
  small.next = largeHead.next;
  return smallHead.next;
};

解法二

双指针

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function(head, x) {
  if (!head) {
    return head;
  }
  let dummyNode = new ListNode(0, head);
  let slow = dummyNode;
  let fast = head.next;
  while (fast) {
    if (slow.next.val < x) {
      slow = slow.next;
      head = head.next;
    } else if (fast.val < x) {
      head.next = fast.next;
      fast.next = slow.next;
      slow.next = fast;
      slow = slow.next;
    } else {
      head = fast;
    }
    fast = head.next;
  }
  return dummyNode.next;
};

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