LeetCode: 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.

注意结尾指针设为NULL。

/** * 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        if (head == NULL) return NULL;                ListNode* a = new ListNode(0);        ListNode* pSmall = a;        ListNode* b = new ListNode(0);        ListNode* pLarge = b;                while(head != NULL)        {            if (head->val < x)            {                pSmall->next = head;                pSmall = head;                head = head->next;            }            else            {                pLarge->next = head;                pLarge = head;                head = head->next;            }        }                pLarge->next = NULL;        pSmall->next = b->next;        return a->next;    }};