LeetCode | Partition List

题目：

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.

思路：

准备两个指针，一个保存小于target的节点，一个保存大于等于target的节点。循环一次完成整个算法。

代码：

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        ListNode * lt=NULL;        ListNode * p1=NULL;        ListNode * gt=NULL;        ListNode * p2=NULL;                ListNode * p=head;        while(p != NULL)        {            if(p->val < x)            {                if(lt == NULL)                {                    lt = p;                    p1 = lt;                }                else                {                    p1->next = p;                    p1 = p1->next;                }                p = p->next;                p1->next = NULL;            }            else            {                if(gt == NULL)                {                    gt = p;                    p2 = gt;                }                else                {                    p2->next = p;                    p2 = p2->next;                }                p = p->next;                p2->next = NULL;            }        };        if(p1 == NULL)        {            return gt;        }        else        {            p1->next = gt;                    return lt;        }    }};