leetcode -- 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 tox.

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.

[解题思路]

找到第一个比x大的节点，在此节点之前插入所有小于x的节点

在头指针位置先插入一个干扰元素，以保证head永不为空，然后在最后返回的时候删除掉

链表操作，需总结

/**
 * 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
           ListNode p = new ListNode(Integer.MIN_VALUE);
           p.next = head;
           head = p;
           
           ListNode pre = null;
           while(p != null && p.val < x){
               pre = p;
               p = p.next;
           }
           
           if(p != null){
               ListNode cur = pre;
               while(p != null){
                   if(p.val < x){
                        pre.next = p.next;
                        ListNode tmp = cur.next;
                        cur.next = p;
                        p.next = tmp;
                        cur = cur.next;
35                         p = pre;
                   }
                   pre = p;
                   p = p.next;
               }
           }
           return head.next;
    }
}