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

 

题目大意：给定一个链表和值x，将链表划分，使得小于x的结点在大于等于x的结点之前，应该保持两组划分内结点的原始的相对顺序。

思路一：用两个链表，将小于x的结点放到一个链表，将大于等于x的放到另一个链表，最后串起来。

ListNode* partition(ListNode* head, int x) {
        ListNode *node1 = new ListNode(0), *node2 = new ListNode(0), *cur1 = node1, *cur2 = node2;
        while (head) {
            if (head->val < x) {
                cur1->next = head;
                cur1 = cur1->next;
            } else {
                cur2->next = head;
                cur2 = cur2->next;
            }
            head = head->next;
        }
        cur1->next = node2->next;
        cur2->next = nullptr; //不能少
        return node1->next;
    }

思路二：一个链表，但是双指针，慢指针指向小于x的结点组成链表的尾结点，快指针指向大于等于x的结点组成链表的尾结点，注：快指针始终应该不在慢指针的后面。

ListNode* partition(ListNode* head, int x) {
        //slow指向第一个链表的尾结点，fast指向第二个链表的尾结点
        ListNode *res = new ListNode(0), *slow = res, *fast = res, *cur = head, *next;
        while (cur != nullptr) {
            next = cur->next;
            cur->next = nullptr;
            if (cur->val < x) {
                if (slow == fast) { //应该始终保持fast指向的位置>= slow
                    slow->next = cur;
                    slow = cur;
                    fast = cur;
                } else {
                    cur->next = slow->next;
                    slow->next = cur;
                    slow = cur;
                }
            } else {
                fast->next = cur;
                fast = cur;
            }
            cur = next;
        }
        return res->next;
    }