Open guodongxiaren opened 4 years ago
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head = new ListNode(0);
ListNode* node = head;
int carry = 0;
while (l1 || l2) {
int x = l1 ? l1->val: 0;
int y = l2 ? l2->val: 0;
int sum = x+y+carry;
carry = sum/10;
if (l1) {
l1 = l1->next;
}
if (l2) {
l2 = l2->next;
}
ListNode* next_node = new ListNode(sum%10);
node->next = next_node;
node = next_node;
}
if (carry) {
ListNode* next_node = new ListNode(1);
node->next = next_node;
}
return head->next;
}
};
LeetCode 445
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head = new ListNode(0);
ListNode* node = head;
int carry = 0;
while (l1 || l2 || carry) {
int x = l1 ? l1->val: 0;
int y = l2 ? l2->val: 0;
int sum = x+y+carry;
carry = sum/10;
if (l1) {
l1 = l1->next;
}
if (l2) {
l2 = l2->next;
}
ListNode* next_node = new ListNode(sum%10);
node->next = next_node;
node = next_node;
}
return head->next;
}
};
给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。
如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例: