LeetCode之“链表”：Partition List

　　题目链接

　　题目要求：

　　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.

　　该题利用dummy节点能极大方便编程，具体程序如下：

/**
 * 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) {
        if(!head || !head->next)
            return head;
        
        ListNode *dummy = new ListNode(INT_MIN);
        dummy->next = head;
        ListNode *parNode = dummy;
        ListNode *preNode = nullptr, *curNode = nullptr;
        while(parNode && parNode->val < x)
        {
            preNode = parNode;
            parNode = parNode->next;
        }
        
        parNode = preNode;
        if(!parNode || !parNode->next)
        {
            head = dummy->next;
            delete dummy;
            dummy = nullptr;
            return head;
        }
        
        preNode = parNode->next;
        curNode = preNode->next;
        while(curNode)
        {
            if(curNode->val < x)
            {
                ListNode *nextPar = parNode->next, *nextCur = curNode->next;
                parNode->next = curNode;
                curNode->next = nextPar;
                parNode = parNode->next;
                
                curNode = nextCur;
                preNode->next = curNode;
            }
            else
            {
                preNode = preNode->next;
                curNode = curNode->next;
            }
        }
        
        head = dummy->next;
        delete dummy;
        dummy = nullptr;
        
        return head;
    }
};