leetcode86. Partition List

86. 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.

解法

左右两条链表，小于x的放在左链表，否则放在右链表，最后串起来。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode partition(ListNode head, int x) {        if (head == null) {            return null;        }        ListNode leftDummy = new ListNode(0);        ListNode rightDummy = new ListNode(0);        ListNode left = leftDummy;        ListNode right = rightDummy;        while (head != null) {            if (head.val < x) {                left.next = head;                left = head;            } else {                right.next = head;                right = head;            }            head = head.next;        }        right.next = null;        left.next = rightDummy.next;        return leftDummy.next;    }}