题目描述:()

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.

解题思路:

将链表分为两个链表，一个链表上所有节点的值都小于x，另一个链表为大于等于x，然后连接两个链表。

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* partition(ListNode* head, int x) {12         ListNode left(-1);13         ListNode right(-1);14         15         ListNode *pLeft = &left;16         ListNode *pRight = &right;17         for (ListNode *cur = head; cur != NULL; cur = cur->next) {18             if (cur->val < x) {19                 pLeft->next = cur;20                 pLeft = cur;21             } else {22                 pRight->next = cur;23                 pRight = cur;24             }25         }26         27         pLeft->next = right.next;28         pRight->next = NULL;29         30         return left.next;31     }32 };