[LeetCode] 21. Merge Two Sorted Lists 合并有序链表

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

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

和88. Merge Sorted Array类似，数据结构不一样，这里是合并链表。

由于是链表，不能像数组一样有从后面往前写的技巧。

解法1：dummy list，新建一个链表，然后两个链表中从头各取一个元素进行比较，小的写入新链表，直到结束，返回dummy.next。

解法2：recursion，代码简洁，但空间复杂度高O(n)

Java:

public class ListNode { 
    int val; 
    ListNode next; 
    ListNode(int x) { val = x; } 
} 

public class Solution {  
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {  
        ListNode flag = new ListNode(0);  
        ListNode firstflag = flag;  
        while (l1 != null && l2 != null) {
            if(l1.val < l2.val){  
                flag.next = l1;  
                l1 = l1.next;  
            }else {  
                flag.next = l2;  
                l2 = l2.next;  
            }  
            flag = flag.next;  
        }  
        flag.next = l1 != null ? l1 : l2;   
        return firstflag.next;  
    }  
}

Java:

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;
		}
}　

Python: Time: O(n), Space: O(1)

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

    def __repr__(self):
        if self:
            return "{} -> {}".format(self.val, self.next)


class Solution(object):
    def mergeTwoLists(self, l1, l2):
        curr = dummy = ListNode(0)
        while l1 and l2:
            if l1.val < l2.val:
                curr.next = l1
                l1 = l1.next
            else:
                curr.next = l2
                l2 = l2.next
            curr = curr.next
        curr.next = l1 or l2
        return dummy.next

Python: wo

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        head = ListNode(0)
        dummy = head
        while l1 or l2:
            if l1 and l2:
                if l1.val < l2.val:
                    dummy.next = l1
                    dummy = dummy.next # required
                    l1 = l1.next
                else:
                    dummy.next = l2
                    dummy = dummy.next # required
                    l2 = l2.next
            elif l1:
                dummy.next = l1
                break
            elif l2:
                dummy.next = l2
                break
                
        return head.next         

Python: Recursion

class Solution(object):
    def mergeLists(head1, head2):
        temp = None
        if head1 is None:
            return head2

        if head2 is None:
            return head1

        if head1.val <= head2.val:
            temp = head1
            temp.next = mergeLists(head1.next, head2)

        else:
            temp = head2
            temp.next = mergeLists(head1, head2.next)

        return temp　　

Python: Recursive, wo, Time: O(n), Space: O(n)

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: 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　　

Python:  in-place, iteratively   

def mergeTwoLists(self, l1, l2):
    if None in (l1, l2):
        return l1 or l2
    dummy = cur = ListNode(0)
    dummy.next = l1
    while l1 and l2:
        if l1.val < l2.val:
            l1 = l1.next
        else:
            nxt = cur.next
            cur.next = l2
            tmp = l2.next
            l2.next = nxt
            l2 = tmp
        cur = cur.next
    cur.next = l1 or l2
    return dummy.next

C++: Time: O(n), Space: O(1)

/**
 * 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{0};
        auto curr = &dummy;

        while (l1 && l2) {
            if (l1->val <= l2->val) {
                curr->next = l1;
                l1 = l1->next;
            } else {
                curr->next = l2;
                l2 = l2->next;
            }
            curr = curr->next;
        }
        curr->next = l1 ? l1 : l2;

        return dummy.next;
    }
};

C++: Recursive

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(l2->next, l1);
            return l2;
        }
    }
};

类似题目：

[LeetCode] 23. Merge k Sorted Lists 合并k个有序链表

[LeetCode] 88. Merge Sorted Array 合并有序数组

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