题目
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
大意：给出两个从低位到高位排列链表形式的多位数，相加后按原格式输出。
分析
感觉这道题还是比较简单的，除了有的语法有点不熟悉查了下书以外，还都是挺顺利的。
编写时，只要注意进位的判断和退出的条件就可以了，代码如下：
Python
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# Definition for singly-linked list. #class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ answer = ListNode(0) pa = answer p1 = l1 p2 = l2 c = 0 while(True): if p1 != None : val1 = p1.val p1 = p1.next else: val1 = 0 if p2 != None : val2 = p2.val p2 = p2.next else: val2 = 0 sum = val1 + val2 + c pa.val = sum % 10 c = sum / 10 if p1 == None and p2 == None and c == 0: return answer else: pa.next = ListNode(0) pa = pa.next |
C++
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/** * 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) { int sum = 0; int c = 0; int val1,val2; ListNode * answer = new ListNode(0); ListNode * p1 = l1; ListNode * p2 = l2; ListNode * pa = answer; while(p1 != NULL || p2 != NULL || c != 0 ){ if(p1 == NULL){ val1 = 0; }else{ val1 = p1->val; p1 = p1->next; } if(p2 == NULL){ val2 = 0; }else{ val2 = p2->val; p2 = p2->next; } sum = val1 + val2 + c; pa->val = sum % 10; c = sum / 10; if(p1 == NULL && p2 == NULL && c == 0){ break; }else{ pa->next = new ListNode(0); pa = pa->next; } } return answer; } }; |