# 0021. Merge Two Sorted Lists ###### tags: `Leetcode` `Medium` `Linked List` Link: https://leetcode.com/problems/merge-two-sorted-lists/ ## 思路 ### 思路一 递归 ### 思路二 当 l1 和 l2 都不是空链表时,判断 l1 和 l2 哪一个链表的头节点的值更小,将较小值的节点添加到结果里,当一个节点被添加到结果里之后,将对应链表中的节点向后移一位。 ## Code ### 思路一 ```c= /** * 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* l1, ListNode* l2) { if(l1 == nullptr){ return l2; } else if(l2 == nullptr){ return l1; } else if(l1->val < l2->val){ l1->next = mergeTwoLists(l1->next, l2); return l1; } else{ l2->next = mergeTwoLists(l1, l2->next); return l2; } } }; ``` ### 思路二 ```c= /** * 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* l1, ListNode* l2) { if(l1 == nullptr){ return l2; } if(l2 == nullptr){ return l1; } ListNode* head = new ListNode; ListNode* prev = head; while(l1!=nullptr && l2!=nullptr){ if(l1->val < l2->val){ prev->next = l1; l1 = l1->next; } else{ prev->next = l2; l2 = l2->next; } prev = prev->next; } prev->next = l1 == nullptr?l2:l1; return head->next; } }; ``` ## Result ### 思路一 ### 思路二 Runtime: 8 ms, faster than **70.36%** of C++ online submissions for Merge Two Sorted Lists. Memory Usage: 14.8 MB, less than **73.95%** of C++ online submissions for Merge Two Sorted Lists.