Leetcode[2] Add Two Numbers

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本文介绍了一种算法问题的解决方案：两个非负整数以链表形式表示，数字以逆序存储，每个节点包含一个数字。通过提供的Java和C++代码示例，展示了如何将这两个数字相加并返回结果作为链表。

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You are given two linkedlists representing two non-negative numbers. The digits are stored in reverseorder and each of their nodes contain a single digit. Add the two numbers andreturn it as a linked list.

Input: (2 -> 4-> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

JAVA版

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode  cur = dummy;
        int carry = 0;
        while(l1 != null || l2 != null || carry != 0){
            int data1 = 0;
            int data2 = 0;
            if(l1 != null) {
                data1 = l1.val;
                l1 = l1.next;
            } 
            if(l2 != null){
                data2 = l2.val;
                l2 = l2.next;
            }
            int tempRes = data1 + data2 +carry;
            carry = tempRes / 10;
            ListNode tail = new ListNode(tempRes % 10);
            cur.next = tail;
            cur = cur.next;
            
        }
        return dummy.next;
    }
}

C++版

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode dummy(0);
        ListNode* prev = &dummy;
        int carry = 0;
        while(l1 != NULL || l2 != NULL || carry != 0)
        {
            int a = 0;
            int b = 0;
            
            if(l1 != NULL) a = l1->val, l1 = l1->next;
            if(l2 != NULL) b = l2->val, l2 = l2->next;
            int sum = a+b+carry;
            prev->next = new ListNode(sum%10);
            prev = prev->next;
            carry = sum/10;
        }
        
        return dummy.next;
    }
};