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.

Solution: 建立两个链表，一个存值比x小的，一个存值比x大的，然后将两个链表连起来。

要将值大的那个表的最后一个数的next = null。

在建链表的时候我采用了头指针的方式。（-1那个指针）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(head == null) return head;
        ListNode small = new ListNode(-1);
        ListNode large = new ListNode(-1);
        ListNode scur = small;
        ListNode lcur = large;
        while(head != null){
            if(head.val < x){
                scur.next = head;
                scur = head;
            }else{
                lcur.next = head;
                lcur = head;
            }
            head = head.next;
        }
        scur.next = large.next;
        lcur.next = null;
        return small.next;
    }
}

 第二遍： 就在原来链表上修改。 将大于target的数字放到原来list的tail处。

public class Solution {
    public ListNode partition(ListNode head, int x) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ListNode header = new ListNode(-1);
        header.next = head;
        int len = 0;
        ListNode cur = header, tail = null;
        while(cur.next != null){
            cur = cur.next;
            len ++;
        }
        tail = cur;
        cur = header;
        int i = 0;
        while(i < len){
            if(cur.next.val >= x){
                tail.next = cur.next;
                cur.next = cur.next.next;
                tail = tail.next;
                tail.next = null;
            }else{
                cur = cur.next;
            }
            i ++;
        }
        return header.next;
    }
}