算法第二周Add Two Numbers[medium]

Add Two Numbers [Medium]

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Description

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

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

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Analysis

这个问题主要考察的是链表，通过链表来实现多位数的加法，需要注意的有以下几个问题：

• 使用链表指针时候要格外的注意，注意其为空的情况以及next指针的问题
• 再考虑加法时要考虑进位的问题，我使用了一个局部变量carry保存进位
• 多位数加法还要考虑两个加数的位数问题，我一开始是将这些情况分别加以讨论
• 特别要注意最后的结果是否需要进位

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Solution

/** * 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* result = NULL;        ListNode* p = l1;        ListNode* q = l2;        ListNode* temp = NULL;        int carry = 0;        while (p != NULL && q != NULL) {            int r = p->val + q->val+carry;            if ( r >= 10) {                carry = 1;                r = r%10;            } else {                carry = 0;            }            if (result == NULL) {                result = new ListNode(r);                temp = result;            } else {                temp->next = new ListNode(r);                temp = temp->next;            }            p = p->next;            q = q->next;        }        if (p == NULL) {           while (q != NULL) {               int r = carry + q->val;               if ( r >= 10) {                   carry = 1;                   r = r%10;               } else {                   carry = 0;               }               if (result == NULL) {                   result = new ListNode(r);                   temp = result;               } else {                   temp->next = new ListNode(r);                   temp = temp->next;               }            q = q->next;           }          }        if (q == NULL) {            while (p != NULL) {               int r = carry + p->val;               if ( r >= 10) {                   carry = 1;                   r = r%10;               } else {                   carry = 0;               }               if (result == NULL) {                   result = new ListNode(r);                   temp = result;               } else {                   temp->next = new ListNode(r);                   temp = temp->next;               }            p = p->next;           }         }        if (p == NULL && q == NULL&& carry != 0) {            temp->next = new ListNode(1);        }        return result;    }};

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Discussion

上述解法通过了，但我发现我的代码特别的繁琐，有很多相同的代码，我就在想是否可以将某些情况合并而减少代码的重复。
经过改进：

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* result = NULL;        ListNode* p = l1;        ListNode* q = l2;        ListNode* temp = NULL;        int carry = 0;        while (p != NULL || q != NULL) {            int x = (p == NULL) ? 0 : p->val;            int y = (q == NULL) ? 0 : q->val;            int r = x + y+carry;            if ( r >= 10) {                carry = 1;                r = r%10;            } else {                carry = 0;            }            if (result == NULL) {                result = new ListNode(r);                temp = result;            } else {                temp->next = new ListNode(r);                temp = temp->next;            }            if (p != NULL ) p = p->next;            if (q != NULL ) q = q->next;        }        if (carry != 0) {            temp->next = new ListNode(1);        }        return result;    }};