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LeetCode #2 Add Two Numbers
题目
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
翻译: 给出两个非空链表表示的两个非负整数。返回值为一个链表表示的两个数的和。
解答
这一题没有什么难度的，直接计算加法即可，需要注意的是进位和空节点的处理。
头节点不为空
/**
* 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) {
int carry = 0;
ListNode *head = nullptr;
ListNode *tail = nullptr;
while ((l1 != nullptr) || (l2 != nullptr)) {
int left = 0;
int right = 0;
if (l1 != nullptr) {
left = l1->val;
l1 = l1->next;
}
if (l2 != nullptr) {
right = l2->val;
l2 = l2->next;
}
int sum = (left + right + carry);
carry = sum / 10;
if (head == nullptr) {
tail = new ListNode(sum % 10);
head = tail;
} else {
tail->next = new ListNode(sum % 10);
tail = tail->next;
}
}
if (carry != 0) {
tail->next = new ListNode(carry);
}
return head;
}
};头节点为空
/**
* 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) {
int carry = 0;
ListNode head(0);
ListNode *tail = &head;
while ((l1 != nullptr) || (l2 != nullptr)) {
int left = (l1 != nullptr) ? l1->val : 0;
int right = (l2 != nullptr) ? l2->val : 0;
if (l1 != nullptr) {
l1 = l1->next;
}
if (l2 != nullptr) {
l2 = l2->next;
}
int sum = (left + right + carry);
carry = sum / 10;
tail->next = new ListNode(sum % 10);
tail = tail->next;
}
if (carry != 0) {
tail->next = new ListNode(carry);
}
return head.next;
}
};
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