445. Add Two Numbers II 两个数字相加 2

一、题目解释

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

我们可以利用栈来保存所有的元素，然后利用栈的后进先出的特点就可以从后往前取数字了，我们首先遍历两个链表，将所有数字分别压入两个栈s1和s2中，我们建立一个值为0的res节点，然后开始循环，如果栈不为空，则将栈顶数字加入sum中，然后将res节点值赋为sum%10，然后新建一个进位节点head，赋值为sum/10，如果没有进位，那么就是0，然后我们head后面连上res，将res指向head，这样循环退出后，我们只要看res的值是否为0，为0返回res->next，不为0则返回res即可。

二、代码

addTwoNumber.hh

#include<iostream>#include<stack>using namespace std;struct ListNode{int val;ListNode *next;ListNode(int x) : val(x),next(NULL){}};class twoNumber{public:ListNode* addTwoNumbers(ListNode* l1,ListNode* l2){stack<int> s1,s2;while(l1){s1.push(l1->val);l1 = l1->next;}while(l2){s2.push(l2->val);l2 = l2->next;}ListNode* res = new ListNode(0);int sum = 0;while(!s1.empty() || !s2.empty()){if(!s1.empty()) {sum += s1.top(); s1.pop();}if(!s2.empty()) {sum += s2.top(); s2.pop();}res->val = sum % 10;ListNode* head = new ListNode(sum/10);head->next = res;res = head;sum = sum/10;}return res->val == 0 ? res->next : res;}};

测试程序：

#include<iostream>#include "addTwoNumber.hh"using namespace std;int main(){ListNode* l1 = new ListNode(3);ListNode* l2 = new ListNode(5);l1->next = new ListNode(4);l2->next = new ListNode(6);twoNumber twoNum;ListNode* result = twoNum.addTwoNumbers(l1,l2);cout << result->val << result->next->val << endl;return 0;}

参考 http://www.cnblogs.com/grandyang/p/6216480.html