86. Partition List

题目描述

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

方法思路

先来一个错误的代码：
下面的第一块代码是不会通过的原因在于链表没有截止于null，分配了太多的内存
Status: Memory Limit Exceeded

class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head == null || head.next ==null)
            return head;
        ListNode smallHead = new ListNode(0), smallHelp = smallHead;
        ListNode xHead = new ListNode(0), xHelp = xHead, cur = head;
        
        while(cur != null){
            if(cur.val < x){
                smallHelp.next = cur;
                smallHelp = smallHelp.next;
            }else{
                xHelp.next = cur;
                xHelp = xHelp.next;
            }
            cur = cur.next;
        }
        if(smallHead.next == null)
            return xHead.next;
        else if(xHead.next == null)
            return smallHead.next;
        
        smallHelp.next = xHead.next;
        
        return smallHead.next;
    }
}

同时相比于参考方法，我的变量命名太过复杂
关键在于： after.next = null;如果不加这一句，就会出现：MemoryLimit Exceeded

class Solution {
    //Runtime: 0 ms, faster than 100.00% 
    //
    public ListNode partition(ListNode head, int x) {

        // before and after are the two pointers used to create the two list
        // before_head and after_head are used to save the heads of the two lists.
        // All of these are initialized with the dummy nodes created.
        ListNode before_head = new ListNode(0);
        ListNode before = before_head;
        ListNode after_head = new ListNode(0);
        ListNode after = after_head;

        while (head != null) {

            // If the original list node is lesser than the given x,
            // assign it to the before list.
            if (head.val < x) {
                before.next = head;
                before = before.next;
            } else {
                // If the original list node is greater or equal to the given x,
                // assign it to the after list.
                after.next = head;
                after = after.next;
            }

            // move ahead in the original list
            head = head.next;
        }

        // Last node of "after" list would also be ending node of the reformed list
        after.next = null;//如果不加这一句，就会出现：MemoryLimit Exceeded

        // Once all the nodes are correctly assigned to the two lists,
        // combine them to form a single list which would be returned.
        before.next = after_head.next;

        return before_head.next;
    }
}