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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

 

AC code:

/**
 * 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* smallHead = new ListNode(0);
        ListNode* bigHead = new ListNode(0);
        ListNode* small = smallHead;
        ListNode* big = bigHead;
        while (head != NULL) {
            ListNode* temp = new ListNode(head->val);
            if (head->val < x) {
                small->next = temp;
                small = small->next;
            } else {
                big->next = temp;
                big = big->next;
            }
            head = head->next;
        }
        small->next = bigHead->next;
        return smallHead->next;
    }
};

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Partition List.

 

这一行代码，写的时候没有注意到：

ListNode* temp = new ListNode(head->val);