leetcode_效率题解_[python/C++]_21. Merge Two Sorted Lists(合并2个有序链表）

相关题解：

leetcode_效率题解_23. Merge k Sorted Lists(合并k个有序链表）

题目链接

【题目】

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.



【分析】

解法1：

class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
ListNode* new_list = NULL;
if(l1->val < l2->val)
{
new_list = l1;
l1 = l1->next;

}
else
{
new_list = l2;
l2 = l2->next;
}
ListNode * pre = new_list;
while( l1 != NULL  && l2 != NULL ){
if( l1->val < l2->val ){
pre->next = l1;
l1 = l1->next;
}
else{
pre->next = l2;
l2 = l2->next;
}
pre = pre->next;
}

if( l1 ) pre->next = l1;
else pre->next = l2;
return new_list;

}
};

解法2：

前面加一个无关的节点，不用去比较两个链表第一个的大小

class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(!l1 && !l2) return NULL;
ListNode * new_list = new ListNode(0);
ListNode * pre = new_list;
while( l1 != NULL  && l2 != NULL ){
if( l1->val < l2->val ){
pre->next = l1;
l1 = l1->next;
}
else{
pre->next = l2;
l2 = l2->next;
}
pre = pre->next;
}
if( l1 )  pre->next = l1;
else  pre->next = l2;
return new_list->next;

}
};

解法3：

递归，leetcode上效率比较高

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;
}
}
};

python：

解法1：非递归：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if l1 is None and l2 is None:
return None
new_list = ListNode(0)
pre = new_list
while l1 is not None and l2 is not None:
if l1.val < l2.val:
pre.next = l1
l1 = l1.next
else:
pre.next = l2
l2 = l2.next
pre = pre.next
if l1 is not None:
pre.next = l1
else:
pre.next = l2
return new_list.next

解法2：递归：

def mergeTwoLists(self, l1, l2):
if not l1:
return l2
elif not l2:
return l1
else:
if l1.val <= l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2

解法3：

discuss上的一种解法，效率很高，但是我是不可能想到的

class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
new_list = ListNode(0)
new_list.next = l1
prev, head_of_1, head_of_2 = new_list, l1, l2

while head_of_2:
if not head_of_1:
prev.next = head_of_2
break
if head_of_1.val > head_of_2.val:
temp = head_of_2
head_of_2 = head_of_2.next
prev.next = temp
temp.next = head_of_1
prev = prev.next
else:
head_of_1, prev = head_of_1.next, head_of_1
return new_list.next