1.题目要求:
给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。
你可以假设除了数字 0 之外，这两个数字都不会以零开头。
示例1：
输入：l1 = [7,2,4,3], l2 = [5,6,4]
输出：[7,8,0,7]
示例2：
输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[8,0,7]
示例3：
输入：l1 = [0], l2 = [0]
输出：[0]
提示：
链表的长度范围为 [1, 100]
0 <= node.val <= 9
输入数据保证链表代表的数字无前导 0
2.做题步骤:
1.遍历两个链表，得到它们的节点数:
struct ListNode* cur1 = l1;
struct ListNode* cur2 = l2;
int count_1 = 0;
int count_2 = 0;
while(cur1)
{
count_1++;
cur1 = cur1->next;
}
while(cur2)
{
count_2++;
cur2 = cur2->next;
}
2.然后创造两个数组，把两个链表的值放入两个数组中:
int* number1 = (int*)malloc(sizeof(int) * count_1);
int* number2 = (int*)malloc(sizeof(int) * count_2);
cur1 = l1;
cur2 = l2;
int i_1 = 0;
int i_2 = 0;
while(cur1)
{
number1[i_1] = cur1->val;
cur1 = cur1->next;
i_1++;
}
while(cur2){
number2[i_2] = cur2->val;
i_2++;
cur2 = cur2->next;
}
3.然后再判断两个数组的长度的，把两个数组相加的值，放入长的数组:
if(count_1 > count_2)
{
int i = 0;
for(i = 0;i < count_1;i++)
{
if(count_1 - i == count_2)
{
break;
}
}
int j = 0;
for(j = 0;j < count_2;j++)
{
number1[i] += number2[j];
i++;
}
i = count_1 - 1;
for(i = count_1 - 1;i > 0;i--){
if(number1[i] > 9)
{
number1[i - 1] += 1;
number1[i] %= 10;
}
}
if(number1[0] > 9)
{
struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
head->next = l1;
l1 = head;
cur1 = l1;
cur1->val = number1[0] / 10;
cur1 = cur1->next;
cur1->val = number1[0] % 10;
cur1 = cur1->next;
for(i = 1;i < count_1;i++)
{
cur1->val = number1[i];
cur1 = cur1->next;
}
return l1;
}else{
cur1 = l1;
for(i = 0;i < count_1;i++)
{
cur1->val = number1[i];
cur1 = cur1->next;
}
return l1;
}
}else{
int i = 0;
for(i = 0;i < count_2;i++)
{
if(count_2 - i == count_1)
{
break;
}
}
int j = 0;
for(j = 0;j < count_1;j++)
{
number2[i] += number1[j];
i++;
}
i = count_2 - 1;
for(i = count_2 - 1;i > 0;i--)
{
if(number2[i] > 9)
{
number2[i - 1] += 1;
number2[i] %= 10;
}
}
if(number2[0] > 9)
{
struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
head->next = l2;
l2 = head;
cur2 = l2;
cur2->val = number2[0] / 10;
cur2 = cur2->next;
cur2->val = number2[0] % 10;
cur2 = cur2->next;
for(i = 1;i < count_2;i++)
{
cur2->val = number2[i];
cur2 = cur2->next;
}
return l2;
}else{
cur2 = l2;
for(i = 0;i < count_2;i++)
{
cur2->val = number2[i];
cur2 = cur2->next;
}
return l2;
}
}
全部代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
struct ListNode* cur1 = l1;
struct ListNode* cur2 = l2;
int count_1 = 0;
int count_2 = 0;
while(cur1)
{
count_1++;
cur1 = cur1->next;
}
while(cur2)
{
count_2++;
cur2 = cur2->next;
}
int* number1 = (int*)malloc(sizeof(int) * count_1);
int* number2 = (int*)malloc(sizeof(int) * count_2);
cur1 = l1;
cur2 = l2;
int i_1 = 0;
int i_2 = 0;
while(cur1)
{
number1[i_1] = cur1->val;
cur1 = cur1->next;
i_1++;
}
while(cur2){
number2[i_2] = cur2->val;
i_2++;
cur2 = cur2->next;
}
if(count_1 > count_2)
{
int i = 0;
for(i = 0;i < count_1;i++)
{
if(count_1 - i == count_2)
{
break;
}
}
int j = 0;
for(j = 0;j < count_2;j++)
{
number1[i] += number2[j];
i++;
}
i = count_1 - 1;
for(i = count_1 - 1;i > 0;i--){
if(number1[i] > 9)
{
number1[i - 1] += 1;
number1[i] %= 10;
}
}
if(number1[0] > 9)
{
struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
head->next = l1;
l1 = head;
cur1 = l1;
cur1->val = number1[0] / 10;
cur1 = cur1->next;
cur1->val = number1[0] % 10;
cur1 = cur1->next;
for(i = 1;i < count_1;i++)
{
cur1->val = number1[i];
cur1 = cur1->next;
}
return l1;
}else{
cur1 = l1;
for(i = 0;i < count_1;i++)
{
cur1->val = number1[i];
cur1 = cur1->next;
}
return l1;
}
}else{
int i = 0;
for(i = 0;i < count_2;i++)
{
if(count_2 - i == count_1)
{
break;
}
}
int j = 0;
for(j = 0;j < count_1;j++)
{
number2[i] += number1[j];
i++;
}
i = count_2 - 1;
for(i = count_2 - 1;i > 0;i--)
{
if(number2[i] > 9)
{
number2[i - 1] += 1;
number2[i] %= 10;
}
}
if(number2[0] > 9)
{
struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
head->next = l2;
l2 = head;
cur2 = l2;
cur2->val = number2[0] / 10;
cur2 = cur2->next;
cur2->val = number2[0] % 10;
cur2 = cur2->next;
for(i = 1;i < count_2;i++)
{
cur2->val = number2[i];
cur2 = cur2->next;
}
return l2;
}else{
cur2 = l2;
for(i = 0;i < count_2;i++)
{
cur2->val = number2[i];
cur2 = cur2->next;
}
return l2;
}
}
}
好了，这就是我的代码，大家如果觉得好的话，就给个免费的赞吧，谢谢各位大佬们了 ^ _ ^ .
