Leetcode_86_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.

题目：给一个单项链表和一个x，重新排序分成两部分，前半部分比x小，后半部分比x大，但是不允许改变链表原本的顺序。
思路：建两个空链表，一个放大的，一个放小的，然后最后把他们连起来就行了。
坑：连接的时候要把greater指针的next清空然后再用，less的next指向greater的头指针，否则容易出现死循环。

/** * 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) {        ListNode *ans = getAns(head, x);        return ans;    }    ListNode* getAns(ListNode* head, int x)    {        ListNode *less_h = NULL;        ListNode *greater_h = NULL;        ListNode *less = NULL;        ListNode *greater = NULL;        ListNode *iter = head;        while(iter!=NULL)        {            if(iter->val<x)            {                if(less_h == NULL)                {                    less_h = iter;                    less = iter;                }                else                {                    less->next = iter;                    less = less->next;                }            }            else            {                if(greater_h == NULL)                {                    greater_h = iter;                    greater = iter;                }                else                {                    greater->next = iter;                    greater = greater->next;                }            }            iter = iter->next;        }        if(greater!=NULL) greater->next = NULL;        if(less_h == NULL) return greater_h;        else         {            less->next = greater_h;            return less_h;        }    }};