LeetCode#445 Add Two Numbers II (week15)

week15

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.

原题地址：https://leetcode.com/problems/add-two-numbers-ii/description/

解析

题目要求以链表形式给出两个数，求这两个数的和，同样以链表形式给出结果。
大致思路是将表示同一个数的每一位的链表中的元素存入栈中，将两个加数的相同位从各自的栈中取出，相加，记录和以及进位，进位作为下一位的数相加时的一个输入。
详细过程参考下面代码及注释

代码

class Solution {public:    ListNode * addTwoNumbers(ListNode* l1, ListNode* l2) {        /* 用来记录两个数的每一位 */        stack<int> s1, s2;        /* 保存相加结果的每一位 */        stack<int> s;        ListNode* tmp;        tmp = l1;        /* 将链表中的元素转移到栈中，方便后面相加 */        while (tmp != NULL) {            s1.push(tmp->val);            tmp = tmp->next;        }        tmp = l2;        while (tmp != NULL) {            s2.push(tmp->val);            tmp = tmp->next;        }        int length1 = s1.size();        int length2 = s2.size();        int num1, num2;        int carry = 0;        /* 第一个数的长度小于或等于第二个数的情况 */        if (length1 <= length2) {            /* 将两个数的相同位相加 */            for (int i = 0; i < length1; ++i) {                num1 = s1.top();                num2 = s2.top();                /* 除了加上当前位置两个数对应的位的数，还要加上上一位的进位 */                s.push((num1 + num2 + carry) % 10);                carry = (num1 + num2 + carry) / 10;                s1.pop();                s2.pop();            }            for (int i = 0; i < length2 - length1; ++i) {                num2 = s2.top();                s.push((num2 + carry) % 10);                carry = (num2 + carry) / 10;                s2.pop();            }        }        /* 第一个数的长度大于第二个数的情况 */        else {            for (int i = 0; i < length2; ++i) {                num1 = s1.top();                num2 = s2.top();                s.push((num1 + num2 + carry) % 10);                carry = (num1 + num2 + carry) / 10;                s1.pop();                s2.pop();            }            for (int i = 0; i < length1 - length2; ++i) {                num1 = s1.top();                s.push((num1 + carry) % 10);                carry = (num1 + carry) / 10;                s1.pop();            }        }        /* 如果最高位有进位，结果需要新增一位数 */        if (carry == 1) {            s.push(carry);        }        ListNode* rel;        ListNode* temp;        /* 需要结果第一位的指针，将第一位特殊处理 */        if (!s.empty()) {            rel = new ListNode(s.top());            temp = rel;            s.pop();        }        /* 将栈中保存的结果的每一位链接成新的链表 */        while (!s.empty()) {            temp->next = new ListNode(s.top());            temp = temp->next;            s.pop();        }        return rel;    }};