You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
首先使用链表实现等长无进位的求和，即实现 (1 —> 2 -> 3) + (1 -> 2 -> 3)=（2 -> 3 -> 6）
1 #include <iostream>
2
3 using namespace std;
4
5 struct ListNode {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11
12 class Solution {
13 public:
14 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
15
16 int val1 = 0, val2 = 0;
17
18 ListNode *l = NULL;
19 ListNode *l_end = NULL;
20 ListNode *l_new = NULL;
21
22 val1 = l1->val;
23 val2 = l2->val;
24 l1 = l1->next;
25 l2 = l2->next;
26 l_new = (ListNode *)new ListNode((val1 + val2) % 10);
27 l_new->next = NULL;
28 l = l_new;
29 l_end = l;
30 while (l1 != NULL || l2 != NULL){
31
32 val1 = l1->val;
33 val2 = l2->val;
34 l1 = l1->next;
35 l2 = l2->next;
36
37 l_new = (ListNode *)new ListNode((val1 + val2)%10);
38 l_new->next = NULL;
39
40 l_end->next = l_new;
41 l_end = l_new;
42 }
43
44 return l;
45
46 }
47 };
48
49 int main()
50 {
51 ListNode *l1, *l2;
52 ListNode *l;
53 l1 = (ListNode *)new ListNode(1);
54 l2 = (ListNode *)new ListNode(1);
55 l1->next = (ListNode *)new ListNode(2);
56 l2->next = (ListNode *)new ListNode(2);
57 l1->next->next = (ListNode *)new ListNode(3);
58 l2->next->next = (ListNode *)new ListNode(3);
59 Solution s;
60 l = s.addTwoNumbers(l1, l2);
61 while (l != NULL){
62 cout << l->val << endl;
63 l = l->next;
64 }
65 while (1);
66 }
运行结果：
然后实现不等长无进位的求和，即实现 (1 —> 2 -> 3) + (1)=（2 -> 2 -> 3）
1 #include <iostream>
2
3 using namespace std;
4
5 struct ListNode {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11
12 class Solution {
13 public:
14 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
15
16 int val1 = 0, val2 = 0;
17
18 ListNode *l = NULL;
19 ListNode *l_end = NULL;
20 ListNode *l_new = NULL;
21
22 val1 = l1->val;
23 val2 = l2->val;
24 l1 = l1->next;
25 l2 = l2->next;
26 l_new = (ListNode *)new ListNode((val1 + val2) % 10);
27 l_new->next = NULL;
28 l = l_new;
29 l_end = l;
30 while (l1 != NULL || l2 != NULL){
31 if (l1 == NULL){
32 val2 = l2->val;
33 l2 = l2->next;
34 val1 = 0;
35 }
36 else if (l2 == NULL){
37 val1 = l1->val;
38 l1 = l1->next;
39 val2 = 0;
40 }
41 else{
42 val1 = l1->val;
43 val2 = l2->val;
44 l1 = l1->next;
45 l2 = l2->next;
46 }
47
48 l_new = (ListNode *)new ListNode((val1 + val2)%10);
49 l_new->next = NULL;
50
51 l_end->next = l_new;
52 l_end = l_new;
53 }
54
55 return l;
56
57 }
58 };
59
60 int main()
61 {
62 ListNode *l1, *l2;
63 ListNode *l;
64 l1 = (ListNode *)new ListNode(1);
65 l2 = (ListNode *)new ListNode(1);
66 l1->next = (ListNode *)new ListNode(2);
67 //l2->next = (ListNode *)new ListNode(2);
68 l1->next->next = (ListNode *)new ListNode(3);
69 //l2->next->next = (ListNode *)new ListNode(3);
70 Solution s;
71 l = s.addTwoNumbers(l1, l2);
72 while (l != NULL){
73 cout << l->val << endl;
74 l = l->next;
75 }
76 while (1);
77 }
运行结果：
最后实现不等长有进位的求和，即实现题目要求（注意最后一位有进位的情况）
1 #include <iostream>
2
3 using namespace std;
4
5 struct ListNode {
6 int val;
7 ListNode *next;
8 ListNode(int x) : val(x), next(NULL) {}
9 };
10
11 class Solution {
12 public:
13 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
14
15 int val1 = 0, val2 = 0, carry = 0;
16
17 ListNode *l = NULL;
18 ListNode *l_end = NULL;
19 ListNode *l_new = NULL;
20
21 val1 = l1->val;
22 val2 = l2->val;
23 l1 = l1->next;
24 l2 = l2->next;
25 l_new = (ListNode *)new ListNode((val1 + val2 + carry) % 10);
26 l_new->next = NULL;
27 l = l_new;
28 carry = (val1 + val2 + carry) / 10;
29 l_end = l;
30 while (l1 != NULL || l2 != NULL){
31 if (l1 == NULL){
32 val2 = l2->val;
33 l2 = l2->next;
34 val1 = 0;
35 }
36 else if (l2 == NULL){
37 val1 = l1->val;
38 l1 = l1->next;
39 val2 = 0;
40 }
41 else{
42 val1 = l1->val;
43 val2 = l2->val;
44 l1 = l1->next;
45 l2 = l2->next;
46 }
47
48 l_new = (ListNode *)new ListNode((val1 + val2 + carry)%10);
49 l_new->next = NULL;
50 carry = (val1 + val2 + carry) / 10;
51
52 l_end->next = l_new;
53 l_end = l_new;
54 }
55 if (carry != 0){
56 l_new = (ListNode *)new ListNode(carry);
57 l_new->next = NULL;
58 l_end->next = l_new;
59 l_end = l_new;
60 }
61 return l;
62
63 }
64 };
65
66 int main()
67 {
68 ListNode *l1, *l2;
69 ListNode *l;
70 l1 = (ListNode *)new ListNode(1);
71 l2 = (ListNode *)new ListNode(1);
72 l1->next = (ListNode *)new ListNode(2);
73 l2->next = (ListNode *)new ListNode(9);
74 l1->next->next = (ListNode *)new ListNode(2);
75 l2->next->next = (ListNode *)new ListNode(7);
76 Solution s;
77 l = s.addTwoNumbers(l1, l2);
78 while (l != NULL){
79 cout << l->val << endl;
80 l = l->next;
81 }
82 while (1);
83 }
运行结果：
因为是一边学C++，一边刷leetcode，所以有什么问题，十分感谢您能指点。
时间： 2024-11-11 12:48:31