[leetcode] Partition List

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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.

https://oj.leetcode.com/problems/partition-list/

 

思路：建立兩個鏈表頭，然後遍曆原鏈表根據大小分別添加到新鏈表頭後邊，最後合并兩個新鏈表。

 

public class Solution {    public ListNode partition(ListNode head, int x) {        if (head == null)            return head;        ListNode head1 = new ListNode(-1);        ListNode head2 = new ListNode(-1);        ListNode p = head;        ListNode p1 = head1;        ListNode p2 = head2;        while (p != null) {            if (p.val < x) {                p1.next = p;                p1 = p1.next;            } else {                p2.next = p;                p2 = p2.next;            }            p = p.next;        }        p1.next = head2.next;        p2.next = null;        return head1.next;    }    public static void main(String[] args) {        ListNode head = ListUtils.makeList(new int[] { 1, 4, 3, 2, 5, 2 });        ListUtils.printList(head);        new Solution().partition(head, 3);        ListUtils.printList(head);    }}

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