guodongxiaren / OJ

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LeetCode 2: 两数相加 #42

Open guodongxiaren opened 4 years ago

guodongxiaren commented 4 years ago

https://leetcode-cn.com/problems/add-two-numbers

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:

输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
guodongxiaren commented 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;
    }

};
guodongxiaren commented 4 years ago

关联题目

LeetCode 445

guodongxiaren commented 4 years ago

good taste!

/**
 * 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;
    }

};