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;
}