Add Two Numbers

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