LeetCode445. Add Two Numbers II

题目

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

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

思路

上一题是最高有效位在链表表尾，现在是最低有效位在表尾，而且要求不将链表结构逆转，所以借助stack或者vector导出链表的数据，才能从最低有效位开始计算，时空复杂度都是0(n)

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) {        ListNode *head = nullptr, *curr = nullptr;        stack<int> temp1, temp2;        while(l1) {            temp1.push(l1->val);            l1 = l1->next;        }        while(l2) {            temp2.push(l2->val);            l2 = l2->next;        }        int sum = 0, carry = 0;        while(!temp1.empty() || !temp2.empty() || carry) {            sum += carry;            if(temp1.size()) {                sum +=  temp1.top();                temp1.pop();            }            if(temp2.size()) {                sum +=  temp2.top();                temp2.pop();            }            //注意先生成的低有效位在表尾            curr = new ListNode(sum % 10);            curr->next = head;            head = curr;            carry = sum / 10;            sum = 0;        }        return head;    }};