LeetCode Add Two Numbers 两个链表数相加

心已赠人 2021-09-11 04:04 348阅读 0赞

## **一、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   **二、代码** ##

两个列表逐个节点相加，遍历过程类似于归并。1、可以以短列表为基础，先将短列表长度遍历完，再单独将长列表多余部分比例；2、以长列表为基础，加以判断短列表长度不够。  
[归并算法][Link 1]  
另一个需要注意：进位问题

/**
     * 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* ll = new ListNode(0);
            ListNode* head = ll;
            while(l1!=NULL || l2!=NULL){   //||以长为限制，&&以短为限制
                int val1 = 0;
                if(l1!=NULL){               //加以判断
                    val1 = l1->val;
                    l1 = l1->next;
                }
                int val2 = 0;
                if(l2!=NULL){                //加以判断
                    val2 = l2->val;
                    l2 = l2->next;
                }
                int temp = val1+val2+carry;
                ll->next = new ListNode(temp%10);
                carry = temp/10;
                ll = ll->next;
            }
            if(carry>=1){
                ll->next = new ListNode(carry);
            }
            return head->next;
        }
    };

[Link 1]: http://blog.csdn.net/qq_16234613/article/details/52679061?locationNum=1