题目放一下:
思路分析:
举例分析一波:
l1和l2是要合并的两个链表,m是最后组合而成的链表
上代码:
首先是C++的:
使用链表:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode dummy(0);
ListNode* tail = &dummy;
while(l1 && l2) {
if (l1->val < l2->val) {
tail->next = l1;
l1 = l1->next;
} else {
tail->next = l2;
l2 = l2->next;
}
tail = tail->next;
}
if (l1) tail->next = l1;
if (l2) tail->next = l2;
return dummy.next;
}
};
如果使用递归的话:
merge(a, b)
= a if b is empty
= b if a is empty
=a[0] + merge(a[1], b) if a[0] < b[0]
=b[0] + merge(a, b[1]) if a[0] > b[0]
使用递归:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode dummy(0);
ListNode* tail = &dummy;
while(l1 && l2) {
if (l1->val < l2->val) {
tail->next = l1;
l1 = l1->next;
} else {
tail->next = l2;
l2 = l2->next;
}
tail = tail->next;
}
if (l1) tail->next = l1;
if (l2) tail->next = l2;
return dummy.next;
}
};
然后是java的:
使用链表
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode l3 = new ListNode(0);
ListNode tail = l3;
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
while ((l1 != null) && (l2 != null)) {
if (l1.val < l2.val) {
tail.next = l1;
l1 = l1.next;
} else {
tail.next = l2;
l2 = l2.next;
}
tail = tail.next;
}
if (l1 == null) {
tail.next = l2;
}
if (l2 == null) {
tail.next = l1;
}
return l3.next;
}
}
使用递归方法:
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;
}
}
}
时间复杂度分析:
使用链表的话
T(n) = 1 + T(n-1)
= O(n)
使用数组的话
T(n) = n + T(n-1)
= O(n^2)
这篇文章的C++版是参照B站的up主花花酱老师的视频写的,java版是我自己写的,有兴趣可以关注花花酱老师,讲的很好~