LeetCode: Partition List

一次过，链表题无难度

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *less, *larger, *p, *q, *ret;
        less = larger = p = q = ret = NULL;
        while (head) {
            if (head->val < x) {
                if (!less) {
                    less = new ListNode(head->val);
                    p = less;
                }
                else {
                    p->next = new ListNode(head->val);
                    p = p->next;
                }
            }
            else {
                if (!larger) {
                    larger = new ListNode(head->val);
                    q = larger;
                }
                else {
                    q->next = new ListNode(head->val);
                    q = q->next;
                }
            }
            head = head->next;
        }
        if (!less) ret = larger;
        if (!larger) ret = less;
        if (less && larger) {
            p->next = larger;
            ret = less;
        }
        return ret;
    }
};

 C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode Partition(ListNode head, int x) {
        ListNode less = null, larger = null, p = null, q = null, ans = null;
        while (head != null) {
            if (head.val < x) {
                if (less == null) {
                    less = new ListNode(head.val);
                    p = less;
                }
                else {
                    p.next = new ListNode(head.val);
                    p = p.next;
                }
            }
            else {
                if (larger == null) {
                    larger = new ListNode(head.val);
                    q = larger;
                }
                else {
                    q.next = new ListNode(head.val);
                    q = q.next;
                }
            }
            head = head.next;
        }
        if (less == null) ans = larger;
        if (larger == null) ans = less;
        if (less != null && larger != null) {
            p.next = larger;
            ans = less;
        }
        return ans;
    }
}