halfrost
0002. Add Two Numbers | LeetCode Cookbook
https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0002.Add-Two-Numbers/
大神， 我有一个类似的解法，为什么提交的时候系统判定 runtime error: out of memory func addTwoNumbers_solution1(l1 *ListNode, l2 *ListNode) *ListNode { var head *ListNode var current *ListNode var previous *ListNode
n1, n2, carry := 0, 0, 0
for i := 0; l1 != nil || l2 != nil || carry != 0; i++ {
if l1 == nil {
n1 = 0
} else {
n1 = l1.Val
l1 = l1.Next
}
if l2 == nil {
n2 = 0
} else {
n2 = l2.Val
l2 = l2.Next
}
current = &ListNode{Val: (n1 + n2 + carry) % 10}
if i == 0 {
head = current
previous = current
}
previous.Next = current
previous = current
carry = (n1 + n2 + carry) / 10
}
return head
}
@xqianwang 大神， 我有一个类似的解法，为什么提交的时候系统判定 runtime error: out of memory func addTwoNumbers_solution1(l1 *ListNode, l2 *ListNode) *ListNode { var head *ListNode var current *ListNode var previous *ListNode
n1, n2, carry := 0, 0, 0
for i := 0; l1 != nil || l2 != nil || carry != 0; i++ { if l1 == nil { n1 = 0 } else { n1 = l1.Val l1 = l1.Next } if l2 == nil { n2 = 0 } else { n2 = l2.Val l2 = l2.Next }
current = &ListNode{Val: (n1 + n2 + carry) % 10} if i == 0 { head = current previous = current } previous.Next = current previous = current carry = (n1 + n2 + carry) / 10}
return head
}
您好，你这个代码的问题在于链表成环了，所以最终导致内存爆掉了。你可以试试这样的测试用例：[]int{1}, []int{1}，两个链表都只有一个数字，你就能发现你 previous.Next = current 这句话就会导致链表成环了。
@hujun2020 测试覆盖率有现成的工具，每次写完自己手动本地跑一下就行。beats 100% 这个工具实现不了，因为是 LeetCode 后端检测 runtime ，然后排名的，并且排名也是动态的。随时会有人超过你。
python打卡
class ListNode:
def __init__(nacho, val = 0, next = None):
nacho.val = val
nacho.next = next
l1List = [ListNode(int(_)) for _ in input("Input: l1 = ").split()]
l2List = [ListNode(int(_)) for _ in input("l2 = ").split()]
for l in [l1List, l2List]:
for index in range(len(l) - 1):
l[index].next = l[index + 1]
l1, l2 = l1List[0], l2List[0]
answer = pa = ListNode((l1.val + l2.val) % 10)
carry = (l1.val + l2.val) // 10
l1, l2 = l1.next, l2.next
while l1 or l2:
val = 0
if l1:
val += l1.val
l1 = l1.next
if l2:
val += l2.val
l2 = l2.next
val += carry
pa.next = ListNode(val % 10)
pa = pa.next
carry = val // 10
if carry:
pa.next = ListNode(carry)
print("Output: ", end = '[')
print(answer.val, end = '')
answer = answer.next
while answer:
print(', ', end = '')
print(answer.val, end = '')
answer = answer.next
print(']')
有点长，实际提交的是中间一段，前面是构造leetcode的数据形式，后面打印是方便自己调试
Java打卡2. 两数相加
/**
* 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 addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = new ListNode();
ListNode node = head;
int carry = 0;
while (carry > 0 || l1 != null || l2 != null) {
int num = carry;
if (l1 != null) {
num += l1.val;
l1 = l1.next;
}
if (l2 != null) {
num += l2.val;
l2 = l2.next;
}
carry = num / 10;
num %= 10;
node.next = new ListNode(num);
node = node.next;
}
return head.next;
}
}
LinkedList<int> Add(LinkedList<int> a, LinkedList<int> b)
{
var af = a.First;
var bf = b.First;
LinkedList<int> r = new(new[] {0});
LinkedListNode<int> rf = r.First;
int carry = 0;
while (af != null || bf != null || carry != 0)
{
int value = (af?.Value ?? 0) + (bf?.Value ?? 0) + carry;
carry = value / 10;
r.AddAfter(rf, value % 10);
rf = rf.Next;
af = af?.Next;
bf = bf?.Next;
}
r.RemoveFirst();
return r;
}
C# 打卡
