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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
解题思路:这题很简单,主要考虑进位问题,每次递增两个列表结点,分情况讨论即可.
#include<iostream> #include<vector> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode*New_head = new ListNode(0); ListNode*Tmp_head = New_head; int Ones = 0; int Tens = 0; ListNode*Pre_node = NULL; while (l1 != NULL && l2 != NULL) { Ones = (l1->val + l2->val + Tmp_head->val) % 10; Tens = (l1->val + l2->val + Tmp_head->val) / 10; Tmp_head->val = Ones; Tmp_head->next = new ListNode(Tens); //根据是否进位确定下一个结点的初始值 l1 = l1->next; l2 = l2->next; Pre_node = Tmp_head; Tmp_head = Tmp_head->next; } if (l1==NULL && l2!=NULL) Pre_node->next = addTwoNumbers(l2, Tmp_head); else if (l2 == NULL&&l1!=NULL) Pre_node->next = addTwoNumbers(l1, Tmp_head); else{ if (Tmp_head->val == 0){ Pre_node->next = NULL; delete Tmp_head; } } return New_head; }
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原文地址:http://blog.csdn.net/li_chihang/article/details/43306631