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Merge Two Sorted Lists⚓︎

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Description⚓︎

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

  • Input: list1 = [1,2,4], list2 = [1,3,4]
  • Output: [1,1,2,3,4,4]

Example 2:

  • Input: list1 = [], list2 = []
  • Output: []

Example 3:

  • Input: list1 = [], list2 = [0]
  • Output: [0]

Constraints:

  • The number of nodes in both lists is in the range [0, 50].
  • -100 <= Node.val <= 100
  • Both list1 and list2 are sorted in non-decreasing order.

Solution⚓︎

Way 1⚓︎

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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* mergeTwoLists(ListNode* list1, ListNode* list2) {
        if (!list1) return list2;
        if (!list2) return list1;
        ListNode* tail = new ListNode();
        ListNode* preHead = tail;
        auto p = list1, q = list2;
        while (p && q) {
            if (p->val <= q->val) {
                tail->next = new ListNode(p->val);
                p = p->next;
            } else {
                tail->next = new ListNode(q->val);
                q = q->next;
            }
            tail = tail->next;
        }
        while (p) {
            tail->next = new ListNode(p->val);
            p = p->next;
            tail = tail->next;
        }
        while (q) {
            tail->next = new ListNode(q->val);
            q = q->next;
            tail = tail->next;
        }
        return preHead->next;
    }
};

Way 2 (In Place)⚓︎

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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* mergeTwoLists(ListNode* list1, ListNode* list2) {
        auto dummy = new ListNode(), tail = dummy;
        while (list1 && list2) {
            if (list1->val < list2->val) {
                tail = tail->next = list1;
                list1 = list1->next;
            } else {
                tail = tail->next = list2;
                list2 = list2->next;
            }
        }
        if (list1) tail->next = list1;
        if (list2) tail->next = list2;

        return dummy->next;
    }
};
  • Time complexity: \(O(m+n)\);
  • Space complexity: \(O(1)\).

Another way of writing (Recommended):

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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* mergeTwoLists(ListNode* list1, ListNode* list2) {
        ListNode preHead;
        auto pre = &preHead;

        while (list1 && list2) {
            if (list1->val < list2->val) {
                pre->next = list1;
                list1 = list1->next;
            } else {
                pre->next = list2;
                list2 = list2->next;
            }
            pre = pre->next;
        }

        pre->next = list1 ? list1 : list2;

        return preHead.next;
    }
};

Way 3 (Recursive)⚓︎

See reference (Chinese).

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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* mergeTwoLists(ListNode* list1, ListNode* list2) {
        if (!list1) return list2;
        if (!list2) return list1;
        if (list1->val < list2->val) {
            list1->next = mergeTwoLists(list1->next, list2);
            return list1;
        } else {
            list2->next = mergeTwoLists(list1, list2->next);
            return list2;
        }
    }
};
  • Time complexity: \(O(m+n)\);
  • Space complexity: \(O(m+n)\).