【[21] 合并两个有序链表】

using namespace std;
#include <algorithm>
#include <array>
#include <bitset>
#include <climits>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <queue>
#include <stack>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
// @lcpr-template-end
// @lc code=start
/* *
 * 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) {}
 * };
 */

//  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* dummy = new ListNode(0);
        ListNode* tail = dummy;
        while (list1 && list2) {
            if (list1->val < list2->val) {
                tail->next = list1;
                list1 = list1->next;
            } else {
                tail->next = list2;
                list2 = list2->next;
            }
            tail = tail->next;
        }
        if (list1) {
            tail->next = list1;
        }
        if (list2) {
            tail->next = list2;
        }
        ListNode* result = dummy->next;
        delete dummy;
        return result;
    }
};

ListNode* dummy = new ListNode(0);
ListNode* tail = dummy;

while (list1 && list2) {
    // 选取较小的节点连接到合并链表的尾部
    if (list1->val < list2->val) {
        tail->next = list1;
        list1 = list1->next;
    } else {
        tail->next = list2;
        list2 = list2->next;
    }
    tail = tail->next;
}

if (list1) {
    tail->next = list1;
}
if (list2) {
    tail->next = list2;
}

ListNode* result = dummy->next;
delete dummy;
return result;