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 ^ _ ^ .