栈的使用以及加法进位的处理

445. Add Two Numbers II

Medium

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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:

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<int> s1;
        stack<int> s2;
        while(l1){
            s1.push(l1->val);
            l1 = l1->next;
        }
        while(l2){
            s2.push(l2->val);
            l2 = l2->next;
        }
        int sum = 0;
        ListNode* Current = new ListNode(0);
        while(!s1.empty() ||!s2.empty()){
            int val1 = 0,val2 = 0;
            if(!s1.empty()){
                val1 = s1.top();
                s1.pop();
            }
            if(!s2.empty()){
                val2 = s2.top();
                s2.pop();
            }
            sum += val1 + val2;
            Current->val = sum % 10;
            ListNode* tmp = new ListNode(sum / 10);
            sum = sum / 10;
            tmp->next = Current;
            Current = tmp;
        }
        if(Current->val == 0){
            return Current->next;
        }
        else{
            return Current;
        }
    }
};