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
解题思路:
这题相当于两个大数相加,只不过这里采用的链表的形式,而不是字符串。
解题时最需注意的是,最后一个节点要考虑会不会进位,over =1时,需要增加一个节点。
实现代码:
#include <iostream>
using namespace std;/**
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*/struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
if(l1 == NULL && l2 == NULL)
return NULL;
ListNode *l3 = new ListNode(-1);
ListNode *tnode = l3;
int over = 0;
while(l1 && l2)
{
int sum = l1->val + l2->val + over;
ListNode *node = new ListNode(sum % 10);
over = sum / 10;
tnode->next = node;
tnode = tnode->next;
l1 = l1->next;
l2 = l2->next;
}
if(l1 == NULL && l2 == NULL && over)//后一个节点,要考虑有没进位
{
ListNode *node = new ListNode(over);
tnode->next = node;
return l3->next;
} ListNode *left = l1;
if(l2)
left = l2;
while(left)
{
int sum = left->val + over;
ListNode *node = new ListNode(sum % 10);
over = sum / 10;
tnode->next = node;
tnode = tnode->next;
left = left->next; }
if(over)//同样,最后一个节点,要考虑有没进位
{
ListNode *node = new ListNode(over);
tnode->next = node;
}
return l3->next; }};
int main(void)
{
return 0;
}