题目
leetcode21. 合并两个有序链表
将两个升序链表合并为一个新的 升序
链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例：
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4
代码
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 (nullptr == l1) {
return l2;
}
if (nullptr == l2) {
return l1;
}
ListNode* head = nullptr;
ListNode* tail = nullptr;
while (l1 && l2) {
if (l1->val <= l2->val) {
if (nullptr == head) {
head = tail = l1;
} else {
tail->next = l1;
tail = tail->next;
}
l1 = l1->next;
} else {
if (nullptr == head) {
head = tail = l2;
} else {
tail->next = l2;
tail = tail->next;
}
l2 = l2->next;
}
if (l1) {
tail->next = l1;
}
if (l2) {
tail->next = l2;
}
}
return head;
}
};
测试
#include <iostream>
#include <vector>
// 创建链表
ListNode* create(std::vector<int> vc) {
ListNode* head = nullptr;
ListNode* tail = nullptr;
auto it = vc.begin();
while (it != vc.end()) {
ListNode* temp = new ListNode();
temp->val = *it;
temp->next = nullptr;
if (nullptr == tail) {
head = temp;
tail = temp;
} else {
tail->next = temp;
tail = tail->next;
}
++it;
}
return head;
}
// 释放链表
void del(ListNode* head) {
while (head) {
auto temp = head->next;
delete head;
head = temp;
}
}
// 打印链表
void print(ListNode* head) {
while (head) {
std::cout << head->val;
head = head->next;
}
std::cout << std::endl;
}
int main() {
std::vector<int> vc1 = {
2, 4, 5, 6, 7};
auto l1 = create(vc1);
print(l1);
std::vector<int> vc2 = {
1, 3, 5};
auto l2 = create(vc2);
print(l2);
Solution s;
auto l = s.mergeTwoLists(l1, l2);
print(l);
del(l);
std::cin.get();
return 0;
}
- 结果
24567
135
12345567