LintCode - Merge Two Sorted List-优快云博客

本文介绍了一种将两个升序排列的链表合并为一个新的升序链表的方法。通过拼接两个链表的节点来实现合并，确保最终链表依然保持升序排列。示例中给出具体的输入输出例子，并提供了 C++ 代码实现。

LintCode - Merge Two Sorted Lists

LintCode - Merge Two Sorted Lists

Web Link

http://www.lintcode.com/en/problem/merge-two-sorted-lists/

Description

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

Code - C++

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
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) {
        // write your code here
        if (l1 == NULL) {
            return l2;
        }
        if (l2 == NULL) {
            return l1;
        }
        ListNode* head = l1->val < l2->val ? l1:l2;
        ListNode* temp = new ListNode(0);
        while (l1 != NULL && l2 != NULL) {
            if (l1->val < l2->val) {
                temp->next = l1;
                temp = l1;
                l1 = l1->next;
            } else {
                temp->next = l2;
                temp = l2;
                l2 = l2->next;
            }
        }
        if (l1 == NULL) {
            temp->next = l2;
        }
        if (l2 == NULL) {
            temp->next = l1;
        }
        return head;
    }
};