链表划分

题目描述：给定一个单链表和数值x，划分链表使得所有小于x的节点排在大于等于x的节点之前。你应该保留两部分内链表节点原有的相对顺序。

样例：给定链表 1->4->3->2->5->2->null，并且 x=3，返回 1->2->2->4->3->5->null

思路:分别申请两个指针left和right,小于x的添加到 left中,大于等于x的添加到right中;最后,将 right链接到left的末端即可

代码如下:

/** * Definition of ListNode * class ListNode { * public: *     int val; *     ListNode *next; *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @param x: an integer     * @return: a ListNode      */    ListNode *partition(ListNode *head, int x) {        // write your code here        if(head == NULL)return head;        ListNode *left = NULL,*right = NULL,*cur = head;        ListNode *r=NULL,*l=NULL;        while(cur){            if(cur->val <x){                if(left == NULL){                    left = cur;                    l = left;                }                else{                    l->next = cur;                    l = cur;                }            }            else{                if(right == NULL){                    right = cur;                    r = right;                }                else{                    r->next = cur;                    r = cur;                }            }            cur = cur->next;        }        if(r != NULL)r->next = NULL;        if(l !=NULL){            l->next = right;            return left;        }        else{            return right;        }            }};