add-two-numbers

题目描述

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

分析：

思路非常简单，利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

java代码

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int carry =0;         ListNode newHead = new ListNode(0);        ListNode p1 = l1, p2 = l2, p3=newHead;         while(p1 != null || p2 != null){            if(p1 != null){                carry += p1.val;                p1 = p1.next;            }             if(p2 != null){                carry += p2.val;                p2 = p2.next;            }             p3.next = new ListNode(carry%10);            p3 = p3.next;            carry /= 10;        }         if(carry==1)             p3.next=new ListNode(1);         return newHead.next;    }}

C++代码

/** * 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* tail = new ListNode(0);        ListNode* ptr = tail;                while(l1 != NULL || l2 != NULL){            int val1 = 0;            if(l1 != NULL){                val1 = l1->val;                l1 = l1->next;            }                        int val2 = 0;            if(l2 != NULL){                val2 = l2->val;                l2 = l2->next;            }                        int tmp = val1 + val2 + carry;            ptr->next = new ListNode(tmp % 10);            carry = tmp / 10;            ptr = ptr->next;        }                if(carry == 1){            ptr->next = new ListNode(1);        }        return tail->next;    }};