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代码

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        if (!head || !(head->next)) return head;
        ListNode *pre = new ListNode(INT_MIN), *ret = pre, *cur = pre;
        pre->next = head;
        while (cur->next) {
            if (cur->next->val >= x)
                cur = cur->next;
            else {
                if (cur == pre) {
                    pre = pre->next;
                    cur = cur->next;
                }
                else {
                    ListNode* tmp = cur->next;
                    cur->next = tmp->next;
                    tmp->next = pre->next;
                    pre->next = tmp;
                    pre = pre->next;
                }
            }
        }
        return ret->next;
    }
};

总结

循规蹈矩的链表基本操作题