Leetcode: Partition List

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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

Analysis: Linked List 惯用套路，Runner Technique(Two Pointers), 一些技巧就是：设置head的前置假节点prev，两个pointer：current和runner都指到这个prev，然后进行判断总是判断 current.next 或者 runner.next. 这样做按照我多次做类似题的经验来说，是最方便省事不容易出错的。这道题一次过。思路就是current 和 runner 一直移动直到找到 current.next >= x 为止，这里就是后面小于x的元素将要插入的位置，current便停在这里，指示这个位置，runner继续往后面寻找，把每一个小于x的元素都插入到current.next 的位置。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode prev = new ListNode(-1);
        prev.next = head;
        ListNode current = prev;
        ListNode runner = prev;
        while (current.next != null && current.next.val < x) {
            current = current.next;
            runner = runner.next;
        }
        while (runner.next != null) {
            if (runner.next.val < x) {
                ListNode temp = runner.next;
                runner.next = runner.next.next;
                temp.next = current.next;
                current.next = temp;
                current = current.next;
            }
            else runner = runner.next;
        }
        return prev.next;
    }
}

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode cur = dummy;
        ListNode runner;
        while (cur.next != null && cur.next.val < x) {
            cur = cur.next;
        }
        runner = cur;
        while (runner.next != null) {
            if (runner.next.val < x) {
                ListNode next = runner.next.next;
                runner.next.next = cur.next;
                cur.next = runner.next;
                cur = cur.next;
                runner.next = next;
            }
            else runner = runner.next;
        }
        return dummy.next;
    }
}