求两个链表结点组成的数之和

如4->4->6 + 2->4 = 4->7->0
#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;
}

每天早上叫醒你的不是闹钟,而是心中的梦~
原文地址:https://www.cnblogs.com/vintion/p/4116876.html