LeetCode 445. 两数相加 II
题目描述
思路分析
加法问题
参考代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2
} else if l2 == nil {
return l1
}
l1, l2 = reverse(l1), reverse(l2)
var pre *ListNode
carry := 0
for l1 != nil || l2 != nil || carry > 0 {
sum := carry
if l1 != nil {
sum += l1.Val
l1 = l1.Next
}
if l2 != nil {
sum += l2.Val
l2 = l2.Next
}
carry = sum / 10
cur := &ListNode{Val: sum % 10}
cur.Next = pre
pre = cur
}
return pre
}
func reverse(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
var pre *ListNode
cur := head
for cur != nil {
node := cur.Next
cur.Next = pre
pre = cur
cur = node
}
return pre
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2
} else if l2 == nil {
return l1
}
stack1, stack2 := make([]int, 0), make([]int, 0)
for l1 != nil {
stack1 = append(stack1, l1.Val)
l1 = l1.Next
}
for l2 != nil {
stack2 = append(stack2, l2.Val)
l2 = l2.Next
}
var pre *ListNode
carry := 0
for len(stack1) > 0 || len(stack2) > 0 || carry > 0 {
sum := carry
if len(stack1) > 0 {
sum += stack1[len(stack1)-1]
stack1 = stack1[:len(stack1)-1]
}
if len(stack2) > 0 {
sum += stack2[len(stack2)-1]
stack2 = stack2[:len(stack2)-1]
}
carry = sum / 10
cur := &ListNode{Val: sum % 10}
cur.Next = pre
pre = cur
}
return pre
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
}
l1 = reverseList(l1);
l2 = reverseList(l2);
ListNode pre = null;
int carry = 0;
while (l1 != null || l2 != null || carry > 0) {
int twoSum = carry;
if (l1 != null) {
twoSum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
twoSum += l2.val;
l2 = l2.next;
}
carry = twoSum / 10;
ListNode cur = new ListNode(twoSum % 10);
cur.next = pre;
pre = cur;
}
return pre;
}
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode pre = null, cur = head;
while (cur != null) {
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
}
本博客所有文章除特别声明外，均采用
CC BY-NC-SA 4.0
许可协议，转载请注明出处！