19.2.23 [LeetCode 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

题意

把链表按前面的全是小于x的，后面的全是大于等于x的顺序重组

题解

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode*less = new ListNode(0),*lessh=less, *greater = new ListNode(0),*greaterh=greater;
        while (head) {
            ListNode*after = head->next;
            if (head->val < x) {
                less->next = head;
                less = less->next;
                less->next = NULL;
            }
            else {
                greater->next = head;
                greater = greater->next;
                greater->next = NULL;
            }
            head = after;
        }
        less->next = greaterh->next;
        return lessh->next;
    }
};

我就总爱各种地方有漏洞……面的时候也差不多……惨