Cracking the coding interview--Q2.4

题目

原文：

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

EXAMPLE

Input: (3 –> 1 –> 5), (5 –> 9 –> 2)

Output: 8 –> 0 –> 8

译文：

你有两个由单链表表示的数。每个结点代表其中的一位数字。数字的存储是逆序的， 也就是说个位位于链表的表头。写一函数使这两个数相加并返回结果，结果也由链表表示。

例子：(3 –> 1 –> 5), (5 –> 9 –> 2)

输出：8 –> 0 –> 8

解答

这道题目并不难，需要注意的有：1.链表为空。2.有进位。3.链表长度不一样。 代码如下：

#include <iostream>
using namespace std;

typedef struct node{
int data;
node *next;
}node;

node* init(int a[], int n){
node *head=NULL, *p;
for(int i=0; i<n; ++i){
node *nd = new node();
nd->data = a[i];
if(i==0){
head = p = nd;
continue;
}
p->next = nd;
p = nd;
}
return head;
}

node* addlink(node *p, node *q){
if(p==NULL) return q;
if(q==NULL) return p;
node *res, *pre=NULL;
int c = 0;
while(p && q){
int t = p->data + q->data + c;
node *r = new node();
r->data = t%10;
if(pre){
pre->next = r;
pre = r;
}
else pre = res = r;
c = t/10;
p = p->next; q = q->next;
}
while(p){
int t = p->data + c;
node *r = new node();
r->data = t%10;
pre->next = r;
pre = r;
c = t/10;
p = p->next;
}
while(q){
int t = q->data + c;
node *r = new node();
r->data = t%10;
pre->next = r;
pre = r;
c = t/10;
q = q->next;
}
if(c>0){//当链表一样长，而又有进位时
node *r = new node();
r->data = c;
pre->next = r;
}
return res;
}

void print(node *head){
while(head){
cout<<head->data<<" ";
head = head->next;
}
cout<<endl;
}

int main(){
int n = 4;
int a[] = {
1, 2, 9, 3
};
int m = 3;
int b[] = {
9, 9, 2
};

node *p = init(a, n);
node *q = init(b, m);
node *res = addlink(p, q);
if(p) print(p);
if(q) print(q);
if(res) print(res);
return 0;
}