Leetcode | 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.

创建两个链表，一个比x小，一个大于等于x。最后把它们连起来。注意把第二个链表最后一个元素置为NULL。不然可能产生死循环。

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode n1(0), n2(0);
        ListNode *p1 = &n1, *p2 = &n2;
        
        while (head != NULL) {
            if (head->val < x) {
                p1->next = head;
                p1 = p1->next;
            } else {
                p2->next = head;
                p2 = p2->next;
            }
            head = head->next;
        }
        p2->next = NULL;
        p1->next = n2.next;
        return n1.next;
    }
};

这里n1和n2不用new，省得最后还要delete.