/**
 * 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}
 */
import { ListNode } from './linked-list-tools';

const partition = function (head, x) {

};

/*
思路：遍历整个链表，把小于x的值连接到一个叫smallHead的哑节点上，把大于等于x的值连接到一个叫largeHead的哑节点上
之后把smallHead整个链表，和largeHead整个链表连接起来就行了
*/
function f1(head, x) {
  let small = new ListNode(0);
  let large = new ListNode(0);
  const smallHead = small;
  const largeHead = large;

  // 遍历整个head表示的列表
  while (head) {
    if (head.val < x) {
      small.next = head;
      small = small.next;
    } else {
      large.next = head;
      large = large.next;
    }
    head = head.next;
  }

  // 连接两个链表
  small.next = largeHead.next;
  large.next = null; // 思考1->2->3->1, x=2 这种情况 samllHead:1->1, largeHead:2->3->1,这个时候3->1是不应该存在的

  return smallHead.next;
}