add two numbers
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.
example:
input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
output: 7 -> 0 -> 8
explanation: 342 + 465 = 807.
code
//
// main.cpp
// 两个数字的加法操作
//
// created by mac on 2019/7/14.
// copyright ? 2019 mac. all rights reserved.
//
#include <iostream>
#include <vector>
#include <list>
#include <algorithm>
using namespace std;
//definition for singly-linked list.
struct listnode {
int val;
listnode *next;
listnode(int x) : val(x), next(null) {}
};
//null是不是0? 是0
class solution {
public:
listnode *addtwonumbers(listnode *l1, listnode *l2) {
listnode prehead(0), *p = &prehead;
int extra = 0;
//如果l1=nullptr 且l2=nullptr 且extra=0 那么这个循环就结束了
while (l1 || l2 || extra) {
if (l1) extra += l1->val, l1 = l1->next;
if (l2) extra += l2->val, l2 = l2->next;
p->next = new listnode(extra % 10);
extra /= 10;
p = p->next;
}
return prehead.next;
}
};
int main(int argc, const char * argv[]) {
// insert code here...
solution so;
listnode *l1=nullptr;
//这个地方的逻辑判断不仅仅限于数字的运算
//if语句不会被执行
if (l1||0) {
cout<<"执行这句话咯,嘿嘿.."<<endl;
}
listnode *l2=nullptr;
for (int i=1; i<4; i++) {
listnode *p=new listnode(i);
p->next=l1;
l1=p;
listnode *q=new listnode(i+1);
q->next=l2;
l2=q;
}
listnode*l3 = so.addtwonumbers(l1, l2);
while (l3!=null) {
cout<<l3->val<<endl;
l3=l3->next;
}
cout<<"+++++++++++++++++++++++"<<endl;
cout<<null<<endl;//null是0
return 0;
}
运行结果
7
5
3
+++++++++++++++++++++++
0
program ended with exit code: 0
参考文献