21. 合并两个有序链表(Merge Two Sorted Lists)E

英文题目

Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists.
Example 1:

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:
```
Input: l1 = [], l2 = []
Output: []
```
Example 3:
```
Input: l1 = [], l2 = [0]
Output: [0]
```
Constraints:
The number of nodes in both lists is in the range [0, 50].
100 <= Node.val <= 100
Both l1 and l2 are sorted in non-decreasing order.

中文题目

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

递归法

终止条件：当两个链表都为空时，表示链表已合并完成
如何递归：判断l1和l2头结点哪个更小，然后较小结点的 next 指针指向其余结点的合并结果

时间复杂度O(m+n)，空间复杂度O(m+n)

# python3: 时间 36 ms, 击败 61.73%; 内存 15.1 MB, 击败 20.34%
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if not l1: return l2
        if not l2: return l1
        if l1.val <= l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

// c++: 时间 8 ms, 击败 61.71%; 内存 14.4 MB, 击败 65.11%
/**
 * 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* list1, ListNode* list2) {
        if (list1 == nullptr) {
            return list2;
        } else if (list2 == nullptr) {
            return list1;
        }
        if (list1->val <= list2->val) {
            list1->next = mergeTwoLists(list1->next, list2);
            return list1;
        } else {
            list2->next = mergeTwoLists(list1, list2->next);
            return list2;
        }
    }
};

// java: 时间 0 ms, 击败 100%; 内存 40.3 MB, 击败 75.16%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        } else if (list2 == null) {
            return list1;
        }
        if (list1.val < list2.val) {
            list1.next = mergeTwoLists(list1.next, list2);
            return list1;
        } else {
            list2.next = mergeTwoLists(list1, list2.next);
            return list2;
        }
    }
}

// go: 时间 0 ms, 击败 100%; 内存 2.4 MB, 击败 27.85%
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
    if list1 == nil {
        return list2
    } else if list2 == nil {
        return list1
    }
    if list1.Val <= list2.Val {
        list1.Next = mergeTwoLists(list1.Next, list2)
        return list1
    } else {
        list2.Next = mergeTwoLists(list1, list2.Next)
        return list2
    }
}

// javascript: 时间 72 ms, 击败 48.39%; 内存 43.6 MB, 击败 5.2%
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} list1
 * @param {ListNode} list2
 * @return {ListNode}
 */
var mergeTwoLists = function(list1, list2) {
    if (list1 == null) {
        return list2;
    } else if (list2 == null) {
        return list1;
    }

    if (list1.val < list2.val) {
        list1.next = mergeTwoLists(list1.next, list2);
        return list1;
    } else {
        list2.next = mergeTwoLists(list1, list2.next);
        return list2;
    }
};

迭代法

基本操作

时间复杂度O(n+m)，空间复杂度O(1)

# python3: 时间 32 ms, 击败 84.58%; 内存 15.1 MB, 击败 20.34%
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode(-1)
        cur = dummy
        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
        
        if l1:
            cur.next = l1
        if l2:
            cur.next = l2
        return dummy.next

// c++: 时间 4 ms, 击败 92.27%; 内存 14.4 MB, 击败 38.37%
/**
 * 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* list1, ListNode* list2) {
        ListNode dummy = ListNode();
        ListNode * cur = &dummy;
        while (list1 != nullptr && list2 != nullptr) {
            if (list1->val <= list2->val) {
                cur->next = list1;
                list1 = list1->next;
            } else {
                cur->next = list2;
                list2 = list2->next;
            }
            cur = cur->next;
        }
        if (list1 != nullptr) {
            cur->next = list1;
        }
        if (list2 != nullptr) {
            cur->next = list2;
        }
        return dummy.next; // 注意此处用.
    }
};

// java: 时间 0 ms, 击败 100%; 内存 40.5 MB, 击败 42.66%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                cur.next = list1;
                list1 = list1.next;
            } else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        if (list1 != null) {
            cur.next = list1;
        }
        if (list2 != null) {
            cur.next = list2;
        }
        return dummy.next;
    }
}

// go: 时间 4 ms, 击败 48.54%; 内存 2.4 MB, 击败 99.90%
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
    dummy := &ListNode{} // 注意此处的初始化
    cur := dummy
    for list1 != nil && list2 != nil {
        if list1.Val <= list2.Val {
            cur.Next = list1
            list1 = list1.Next
        } else {
            cur.Next = list2
            list2 = list2.Next
        }
        cur = cur.Next
    }
    if list1 != nil {
        cur.Next = list1
    }
    if list2 != nil {
        cur.Next = list2
    }
    return dummy.Next
}

// javascript: 时间 68 ms, 击败 69.14%; 内存 43.2 MB, 击败 40.43%
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} list1
 * @param {ListNode} list2
 * @return {ListNode}
 */
var mergeTwoLists = function(list1, list2) {
    let dummy = new ListNode(-1);
    let cur = dummy;
    while (list1 != null && list2 != null) {
        if (list1.val < list2.val) {
            cur.next = list1;
            list1 = list1.next;
        } else {
            cur.next = list2;
            list2 = list2.next;
        }
        cur = cur.next;
    }
    if (list1 != null) {
        cur.next = list1;
    }
    if (list2 != null) {
        cur.next = list2;
    }
    return dummy.next;
};

# 21. 合并两个有序链表(Merge Two Sorted Lists)E

# 英文题目

# 中文题目

# 递归法

# 迭代法

21. 合并两个有序链表(Merge Two Sorted Lists)E

英文题目

中文题目

递归法

迭代法