25. 合并两个排序的链表

常某某的好奇心2024-08-19 15:12

comments: true

difficulty: 简单
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/面试题25. 合并两个排序的链表/README.md

面试题 25. 合并两个排序的链表

题目描述

输入两个递增排序的链表，合并这两个链表并使新链表中的节点仍然是递增排序的。

示例1：

复制代码 
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

限制：

0 <= 链表长度 <= 1000

注意：本题与主站 21 题相同：https://leetcode.cn/problems/merge-two-sorted-lists/

解法

方法一：迭代

我们先创建一个虚拟头结点 dummy，然后创建一个指针 cur 指向 dummy。

接下来，循环比较 l1 和 l2 的值，将较小的值接在 cur 后面，然后将指针向后移动一位。循环结束，将 cur 指向 l1 或者 l2 中剩余的部分。

最后返回 dummy.next 即可。

时间复杂度 O ( m + n ) O(m + n) O(m+n)，空间复杂度 O ( 1 ) O(1) O(1)。其中 m m m 和 n n n 分别为两个链表的长度。

Python3
python 复制代码 
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = cur = ListNode(0)
        while l1 and l2:
            if l1.val <= l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        cur.next = l1 or l2
        return dummy.next
Java
java 复制代码 
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        return dummy.next;
    }
}
C++
cpp 复制代码 
/**
 * 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) {
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy;
        while (l1 && l2) {
            if (l1->val <= l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1 ? l1 : l2;
        return dummy->next;
    }
};
Go
go 复制代码 
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	dummy := &ListNode{}
	cur := dummy
	for l1 != nil && l2 != nil {
		if l1.Val <= l2.Val {
			cur.Next = l1
			l1 = l1.Next
		} else {
			cur.Next = l2
			l2 = l2.Next
		}
		cur = cur.Next
	}
	if l1 == nil {
		cur.Next = l2
	} else {
		cur.Next = l1
	}
	return dummy.Next
}
TypeScript
ts 复制代码 
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function mergeTwoLists(l1: ListNode | null, l2: ListNode | null): ListNode | null {
    const dummy = new ListNode(0);
    let cur = dummy;
    while (l1 && l2) {
        if (l1.val <= l2.val) {
            cur.next = l1;
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
        }
        cur = cur.next;
    }
    cur.next = l1 || l2;
    return dummy.next;
}
Rust
rust 复制代码 
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn merge_two_lists(
        mut l1: Option<Box<ListNode>>,
        mut l2: Option<Box<ListNode>>,
    ) -> Option<Box<ListNode>> {
        match (l1.is_some(), l2.is_some()) {
            (false, false) => None,
            (true, false) => l1,
            (false, true) => l2,
            (true, true) => {
                let mut dummy = Box::new(ListNode::new(0));
                let mut cur = &mut dummy;
                while l1.is_some() && l2.is_some() {
                    cur.next = if l1.as_ref().unwrap().val < l2.as_ref().unwrap().val {
                        let mut res = l1.take();
                        l1 = res.as_mut().unwrap().next.take();
                        res
                    } else {
                        let mut res = l2.take();
                        l2 = res.as_mut().unwrap().next.take();
                        res
                    };
                    cur = cur.next.as_mut().unwrap();
                }
                cur.next = if l1.is_some() { l1.take() } else { l2.take() };
                dummy.next.take()
            }
        }
    }
}
JavaScript
js 复制代码 
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function (l1, l2) {
    const dummy = new ListNode(0);
    let cur = dummy;
    while (l1 && l2) {
        if (l1.val <= l2.val) {
            cur.next = l1;
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
        }
        cur = cur.next;
    }
    cur.next = l1 || l2;
    return dummy.next;
};
C#
cs 复制代码 
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        return dummy.next;
    }
}
Swift
swift 复制代码 
/* public class ListNode {
*     var val: Int
*     var next: ListNode?
*     init(_ val: Int) {
*         self.val = val
*         self.next = nil
*     }
* }
*/

class Solution {
    func mergeTwoLists(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let dummy = ListNode(0)
        var cur: ListNode? = dummy
        var l1 = l1
        var l2 = l2

        while let l1Node = l1, let l2Node = l2 {
            if l1Node.val <= l2Node.val {
                cur?.next = l1Node
                l1 = l1Node.next
            } else {
                cur?.next = l2Node
                l2 = l2Node.next
            }
            cur = cur?.next
        }

        cur?.next = l1 ?? l2

        return dummy.next
    }
}

方法二：递归

我们先判断 l1 和 l2 中有没有一个为空，如果有一个为空，那么我们直接返回另一个链表即可。

接下来，我们比较 l1 和 l2 的值：

  • 如果 l1 的值小于等于 l2 的值，我们递归调用 mergeTwoLists(l1.next, l2)，并将 l1.next 指向返回的链表，然后返回 l1。
  • 如果 l1 的值大于 l2 的值，我们递归调用 mergeTwoLists(l1, l2.next)，并将 l2.next 指向返回的链表，然后返回 l2。

时间复杂度 O ( m + n ) O(m + n) O(m+n)，空间复杂度 O ( m + n ) O(m + n) O(m+n)。其中 m m m 和 n n n 分别为两个链表的长度。

Python3
python 复制代码 
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 is None or l2 is None:
            return l1 or l2
        #回溯时，将产生有序合并
        if l1.val <= l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2
Java
java 复制代码 
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        if (l1.val <= l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}
C++
cpp 复制代码 
/**
 * 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) {
            return l2;
        }
        if (!l2) {
            return l1;
        }
        if (l1->val <= l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};
Go
go 复制代码 
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	}
	if l2 == nil {
		return l1
	}
	if l1.Val <= l2.Val {
		l1.Next = mergeTwoLists(l1.Next, l2)
		return l1
	} else {
		l2.Next = mergeTwoLists(l1, l2.Next)
		return l2
	}
}
TypeScript
ts 复制代码 
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function mergeTwoLists(l1: ListNode | null, l2: ListNode | null): ListNode | null {
    if (l1 == null || l2 == null) {
        return l1 || l2;
    }
    if (l1.val < l2.val) {
        l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    }
    l2.next = mergeTwoLists(l1, l2.next);
    return l2;
}
Rust
rust 复制代码 
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn merge_two_lists(
        l1: Option<Box<ListNode>>,
        l2: Option<Box<ListNode>>,
    ) -> Option<Box<ListNode>> {
        match (l1, l2) {
            (Some(mut n1), Some(mut n2)) => {
                if n1.val < n2.val {
                    n1.next = Self::merge_two_lists(n1.next, Some(n2));
                    Some(n1)
                } else {
                    n2.next = Self::merge_two_lists(Some(n1), n2.next);
                    Some(n2)
                }
            }
            (Some(node), None) => Some(node),
            (None, Some(node)) => Some(node),
            (None, None) => None,
        }
    }
}
JavaScript
js 复制代码 
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function (l1, l2) {
    if (!(l1 && l2)) {
        return l1 || l2;
    }
    if (l1.val < l2.val) {
        l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    } else {
        l2.next = mergeTwoLists(l2.next, l1);
        return l2;
    }
};
C#
cs 复制代码 
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        if (l1.val <= l2.val) {
            l1.next = MergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = MergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}
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