Leetcode:Partition List

Description:

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.

分析：这道题目就是将链表中小于目标值的元素移到前面，链表中所有元素的相对位置都应该保持不变，其实就是一些

最基本的链表操作，但是这里有一个小trick, 因为每次都要往前看一个元素，以确定操作，所以在链表前面增加一个额外元素，

将所有的操作在循环内就搞定了！

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(head == NULL) return head;
        
        ListNode *newh = new ListNode(0);
        newh->next = head;
        
        ListNode *pass= newh, *insertpos=newh, *findnod;
        while(pass->next!=NULL)
        {
            if(pass->next->val>=x)
                pass = pass->next;
            else if(insertpos->next == pass->next)
                 {
                     pass = pass->next;
                     insertpos = insertpos->next;
                 }
                 else{
                     findnod = pass->next;
                     pass->next = findnod->next;
                     //pass = pass->next;
                     
                     findnod->next = insertpos->next;
                     insertpos->next = findnod;
                     insertpos = insertpos->next;
                 }
        }
        
        return newh->next;
    }
};