143. Reorder List

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143. Reorder List

Description

Difficulty: Medium

Related Topics: Linked List, Two Pointers, Stack, Recursion

You are given the head of a singly linked-list. The list can be represented as:

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L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

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L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list’s nodes. Only nodes themselves may be changed.

Example 1:

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Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

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Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]

Constraints:

  • The number of nodes in the list is in the range [1, 5 * 104].
  • 1 <= Node.val <= 1000

Hints/Notes

Solution

Iterative solution:

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
        }
        return slow;
    }

    ListNode* reverseList(ListNode* head) {
        ListNode *cur = head, *prev = nullptr;
        while (cur) {
            ListNode* nxt = cur->next;
            cur->next = prev;
            prev = cur;
            cur = nxt;
        }
        return prev;
    }

    void reorderList(ListNode* head) {
        // get the pointer to the middle node, e.g. 1, 2, 3, return pointer to 2
        ListNode* mid = middleNode(head);
        // two cases:
        // #1:
        // 1 -> 2
        // 3 -> 2
        // #2:
        // 1 -> 2 -> 3
        // 4 -> 3
        mid = reverseList(mid);
        while (mid->next) {
            ListNode* nxt1 = head->next;
            ListNode* nxt2 = mid->next;
            head->next = mid;
            mid->next = nxt1;
            head = nxt1;
            mid = nxt2;
        }
    }
};

Language: C++

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        stack<ListNode*> s;
        ListNode* tmp = head;
        while (tmp) {
            s.push(tmp);
            tmp = tmp->next;
        }
        tmp = head;
        while (tmp != s.top() && tmp->next != s.top()) {
            ListNode* tail = s.top();
            s.pop();
            ListNode* next = tmp->next;
            tmp->next = tail;
            tail->next = next;
            tmp = next;
        }
        if (tmp == s.top()) {
            tmp->next = nullptr;
        } else {
            tmp->next->next = nullptr;
        }
    }
};