86. Partition List

86. 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.

/**
 * 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 == nullptr || head -> next == nullptr){
            return head;
        }
        ListNode* dummyLeft = new ListNode(-1);
        ListNode* pLeft = dummyLeft;
        ListNode* dummyRight = new ListNode(-1);
        ListNode* pRight = dummyRight;
        
        while(head != nullptr){
            if(head -> val < x){
                dummyLeft -> next = head;
                dummyLeft = dummyLeft -> next;
            }
            else{
                dummyRight -> next  = head;
                dummyRight = dummyRight -> next;
            }
            head = head -> next;
        }
        if(dummyLeft == nullptr){
             return pRight -> next;
        }
        dummyLeft -> next = pRight -> next;
        dummyRight -> next = nullptr;//链表的结尾必须nullptr。
        
        return pLeft -> next;
    }
};