86. Partition List (M)

Partition List (M)

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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

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题意

给定一个链表和目标值x，将链表中所有值小于x的结点移到所有值大于等于x的结点的前面，同时不能改变值小于/大于等于x的结点的相对位置。

思路

新建两个头结点分别用来保存原链表中值小于x的结点和值大于等于x的结点，最后将两链表相连即为所求的新链表。

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代码实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        // 建空的头结点便于处理
        ListNode leftHead = new ListNode(0), leftTail = leftHead;
        ListNode rightHead = new ListNode(0), rightTail = rightHead;

        while (head != null) {
            if (head.val < x) {
                leftTail.next = head;
                leftTail = head;
            } else {
                rightTail.next = head;
                rightTail = head;
            }
            head = head.next;
        }

        rightTail.next = null;				// 注意断链，否则可能会形成环
        leftTail.next = rightHead.next;		// 拼接链表

        return leftHead.next;
    }
}