Leetcode 86. Partition List(鏈表劃分)

題目描述

給定鏈表和整型值x，將小於x的結點放到鏈表左邊，大於x的結點放到鏈表右邊。

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.

思路分析：

遍歷鏈表，將小於x的結點插入到左鏈表的尾部，將大於x的結點插入到右鏈表的尾部，然後將左鏈表的尾結點指向右鏈表的頭結點，然後返回左鏈表的頭結點。

代碼實現

/**
 * 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 || head.next == null) {
            return head;
        }
        ListNode leftHead = new ListNode(0);//左鏈表的頭結點
        ListNode leftTail = leftHead;//左鏈表的尾結點
        ListNode rightHead = new ListNode(0);//右鏈表的頭結點
        ListNode rightTail = rightHead;//右鏈表的尾結點
        ListNode p = head;
        while (p != null) {
            if (p.val < x) {
                leftTail.next = p;
                leftTail = leftTail.next;
            } else {
                rightTail.next = p;
                rightTail = rightTail.next;
            }
            p = p.next;
        }
        //將左鏈表的尾結點指向右鏈表的頭結點
        leftTail.next = rightHead.next;
        //將右鏈表的尾結點的下一個結點置爲空
        rightTail.next = null;
        //返回新的頭結點
        return leftHead.next;
    }
}