LeetCode OJ - Add Two Numbers

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

/** * 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) {        if(!l1 || !l2) return NULL;        ListNode *dummy = new ListNode(0);        ListNode *cur = dummy;                bool flag = false;        ListNode *last = l1;         while(l1 || l2) {            int sum = 0;            if(l1) sum += l1->val;            if(l2) sum += l2->val;            if(flag) sum += 1;        //是否有进位            ListNode *node = new ListNode(0);            if(sum >= 10) {                flag = true;                node->val = sum - 10;            } else{                flag = false;                node->val = sum;            }            cur->next = node;            cur = cur->next;                        if(l2) l2 = l2->next;            if(l1) l1 = l1->next;        }        if(flag) {            ListNode *node = new ListNode(1);            cur->next = node;        }         return dummy->next;        }};

链表的操作，下面是一般形式:

dummy(0);dummy.next = head;last = &dummy;cur = dummy.next;while(!cur) {    if(cur->val == val) {    }    last = cur;    cur = cur->next;}return dummy.next;

这里之所以需要dummy节点，就是为了防止链表在操作过程中为空而需要写出多余的逻辑判断，导致代码可读性不好，牺牲一点空间是值得的。同时，双链表的操作，一般采用下面结构：

while(ll && l2) {}while(l1) {}while(l2) {}