Leetcode[21]-Merge Two Sorted Lists

Link: https://leetcode.com/problems/merge-two-sorted-lists/

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

思路：要求合并两个排好序的链表。开始我们初始化头front和尾tail，然后从两个单链表的头部比较两个单链表，两链表同时不为空的条件下递归比较：

• 如果l1的值大于l2的值，就将尾部指向l1，并同步向右移动l1的头指针和tail指针
• 如果l1的值小于l2的值，就将尾部指向l2，并同步向右移动l2的头指针和tail指针

接着，如果l1不为空，则将尾指针的下一个指向l1，如果l2不为空，则将尾指针的下一个指向l2，然后将front右移一位，

最后返回front即可。

Code(c++):

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        if(l1 == NULL) return l2;        if(l2 == NULL) return l1;        ListNode* front = new ListNode(-1);        ListNode* tail = front;        while( l1 && l2){            if(l1->val < l2->val){                tail->next = l1;                l1 = l1->next;                tail = tail->next;            }else {                tail->next = l2;                l2 = l2->next;                tail = tail->next;            }        }        if(l1){            tail->next = l1;        }        if(l2){            tail->next = l2;        }        front = front->next;        return front;    }};