[算法Rust,Go,Python,JS实现)]LeetCode之21-合并两个有序链表

版权声明：未经博主允许不得转载(https://github.com/ai-word) https://blog.csdn.net/BaiHuaXiu123/article/details/88738019

题目

将两个有序链表合并为一个新的有序链表并返回。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

思路
参考链表

单链表相关知识，递归实现

Python实现

class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
result = ListNode(None)
if l1.val >= l2.val:
result = l2
result.next = self.mergeTwoLists(l1,l2.next)
else:
result = l1
result.next = self.mergeTwoLists(l1.next, l2)

return result

Golang实现

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return  l2
}
if l2 == nil {
return l1
}
var result *ListNode
if l1.Val >= l2.Val {
result = l2
result.Next = mergeTwoLists(l1,l2.Next)
} else {
result = l1
result.Next = mergeTwoLists(l1.Next,l2)
}
return result
}

JavaScript实现

var mergeTwoLists = function(l1, l2) {
if (l1 == null) {
return l2
}
if (l2 == null) {
return l1
}
var result = ListNode(null);
if (l1.val >= l2.val) {
result = l2
result.next = mergeTwoLists(l1,l2.next)
} else {
result = l1
result.next = mergeTwoLists(l1.next,l2)
}
return result
};

Rust实现

impl Solution {
pub fn merge_two_lists(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let  l11 = l1.as_ref().unwrap();
let  l22 = l2.as_ref().unwrap();
if let Some(L1) = l1.as_ref() {
println!("{:?}",L1);
} else {
return l2;
}
if let Some(L2) = l2.as_ref() {
println!("{:?}",L2);
} else {
return l1;
}
let mut reslut = Box::new(ListNode::new(-1));

if l11.val >= l22.val {
reslut = l2.unwrap();
let  res = reslut.as_ref();
let mut  n = res.next;
let b: Option<Box<ListNode>> = Solution::merge_two_lists(l1, l22.next);
n = b;
} else {
reslut = l1.unwrap();
let  res = reslut.as_ref();
let mut  n = res.next;
let b: Option<Box<ListNode>> = Solution::merge_two_lists(l11.next,l2);
n = b;
}
return Some(reslut)
}
}

结果

代码下载地址

代码下载地址 https://github.com/ai-word