[LintCode] Merge Two Sorted Lists 混合插入有序链表

 

Merge two sorted (ascending) linked lists and return it as a new sorted list. The new sorted list should be made by splicing together the nodes of the two lists and sorted in ascending order.

Have you met this question in a real interview? 

Yes

Example

Given 1->3->8->11->15->null, 2->null , return 1->2->3->8->11->15->null.

 

LeetCode上的原题，请参见我之前的博客Merge Two Sorted Lists。

 

解法一：

class Solution {
public:
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                cur->next = 1l;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1 ? l1 : l2;
        return dummy->next;
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if (!l1) return l2;
        if (!l2) return l1;
        if (l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};

 

解法三：

class Solution {
public:
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if (!l1) return l2;
        if (!l2) return l1;
        ListNode *head = (l1->val < l2->val) ? l1 : l2;
        ListNode *nonHead = (l1->val < l2->val) ? l2 : l1;
        head->next = mergeTwoLists(head->next, nonHead);
        return head;
    }
};

 

解法四：

class Solution {
public:
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if (!l1 || (l2 && l1->val > l2->val)) swap(l1, l2);
        if (l1) l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    }
};