LeetCode算法题目：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 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.

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分析：

遍历一遍链表，把小于x的都挂到head1后，把大于等于x的都放到head2后，最后再把大于等于的链表挂到小于链表的后面就可以了。

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代码实现：

/** * 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 *head1 = new ListNode(0);        ListNode *head2 = new ListNode(0);        ListNode *h1 = head1;        ListNode *h2 = head2;        while(head)        {            int v = head->val;            if(v < x)            {                head1->next = head;                head1 = head1->next;            } else {                head2->next = head;                head2 = head2->next;            }            head = head->next;        }        head2->next = NULL;        head1->next = h2->next;;        return h1->next;    }};