2021-12-25

两个数相加

给你两个 非空 的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储 一位 数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

【解析】

本题和上一题二进制数相加是一样的，区别只是在于上一题是基于数组的，而本题是基于链表的。把数组操作改为链表操作即可。

【code】

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
   
    ListNode *head = nullptr, *tail = nullptr;
    int carry = 0;
    while (l1 || l2) {
   
        int n1 = l1 ? l1->val: 0;
        int n2 = l2 ? l2->val: 0;
        int sum = n1 + n2 + carry;
        if (!head) {
   
            head = tail = new ListNode(sum % 10);
        } else {
   
            tail->next = new ListNode(sum % 10);
            tail = tail->next;
        }
        carry = sum / 10;
        if (l1) {
   
            l1 = l1->next;
        }
        if (l2) {
   
            l2 = l2->next;
        }
    }
    if (carry > 0) {
   
        tail->next = new ListNode(carry);
    }
    return head;
}

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
   
    ListNode *result = new ListNode(-1);
    ListNode *pre = result;
    ListNode *pa = l1->next, *pb = l2->next;
    int carry = 0;
    while (pa != NULL || pb != NULL){
   
        int av = pa == NULL ? 0 : pa->val;
        int bv = pb == NULL ? 0 : pb->val;
        //cout << "av:" << av<<" bv: "<<bv<<endl;
        ListNode *node = new ListNode((av + bv + carry) % 10);
        carry = (av + bv + carry) / 10;
        pre->next = node;
        pre = pre->next;
        pa = pa == NULL ? NULL : pa->next;
        pb = pb == NULL ? NULL : pb->next;
    }
    if(carry>0)
        pre->next = new ListNode(1);
    pre = result->next;
    delete result;
    return pre;
}

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;

const int inf=0x3f3f3f3f;
const int maxn=1e6+5;

struct ListNode {
   
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {
   }
    ListNode(int x) : val(x), next(nullptr) {
   }
    ListNode(int x, ListNode *next) : val(x), next(next) {
   }
};
void CreateList_L(ListNode* &L,int n){
   //带头结点尾插
    L = (ListNode *)malloc(sizeof(ListNode));
    L->next = NULL;
    ListNode* m = L;
    int e;
	for (int i = n; i > 0;i--){
   
		ListNode* p = (ListNode*)malloc(sizeof(ListNode));
        scanf("%d", &e);
        p->val = e;
        m->next = p;
        m = p;
    }
    m->next = NULL;
}
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
   
    ListNode *result = new ListNode(-1);
    ListNode *pre = result;
    ListNode *pa = l1->next, *pb = l2->next;
    int carry = 0;
    while (pa != NULL || pb != NULL){
   
        int av = pa == NULL ? 0 : pa->val;
        int bv = pb == NULL ? 0 : pb->val;
        cout << "av:" << av<<" bv: "<<bv<<endl;
        ListNode *node = new ListNode((av + bv + carry) % 10);
        carry = (av + bv + carry) / 10;
        pre->next = node;
        pre = pre->next;
        pa = pa == NULL ? NULL : pa->next;
        pb = pb == NULL ? NULL : pb->next;
    }
    if(carry>0)
        pre->next = new ListNode(1);
    pre = result->next;
    delete result;
    return pre;
}
void display_list(ListNode* head){
   
    ListNode* cur = (ListNode*)malloc(sizeof(ListNode));
    cur = head;
    cout << cur->val << " ";
    while ((cur = cur -> next)){
   
        printf("%d ", cur -> val);
    }
    free(cur);
    printf("\n");
}

int main(){
   
    ListNode* l1;
    ListNode* l2;
    CreateList_L(l1, 3);
    CreateList_L(l2, 3);
    ListNode* res=addTwoNumbers(l1, l2);
    display_list(res);
    return 0;
}

测试样例

input:
2 4 3
5 6 4
output:
7 0 8

模拟每一步操作

2 4 3
5 6 4

av:2 bv: 5
av:4 bv: 6
av:3 bv: 4
7 0 8