Leetcode-86. Partition List

解题思路：使用两个节点保存切分后的俩链表。最后再合并到一起。

/**
 * 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 *small = new ListNode(0);
        ListNode *big = new ListNode(0);
        
        ListNode *ps = small,*pb = big;
        ListNode *cur = head;
        while(cur){
            if(cur->val < x){
                ps->next = cur;
                cur = cur->next;
                ps = ps->next;
                ps->next=NULL;
                
            }else{
                pb->next = cur;
                cur = cur->next;
                pb = pb->next;
                pb->next=NULL;                
            }
        }

        ps->next = big->next;
        head = small->next;
        delete small;
        delete big;
        return head;
    }
};