leetcode:2. Add Two Numbers

描述

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

思路

我最终的思路很简单，就是使用轮询链表，并相加。（见代码一）
我第一个思路是将l1，l2全部变成数字，并将这些数字相加，将结果再转变成链表返回，但是由于类型的限制，如果链表太长则没有办法相加。（见代码二）

代码

代码一

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        int carry = 0, sum = 0;        sum = l1->val + l2->val;        struct ListNode* resultList = new ListNode(sum % 10);        struct ListNode* cur = resultList;        carry = sum / 10;        l1 = l1->next;        l2 = l2->next;        while(l1 && l2){            sum = l1->val + l2->val + carry;            carry = sum / 10;            cur-> next = new ListNode(sum % 10);            cur = cur->next;            l1 = l1->next;            l2 = l2->next;        }        while(l1){            sum = l1->val + carry;            carry = sum / 10;            cur-> next = new ListNode(sum % 10);            cur = cur->next;            l1 = l1->next;        }        while(l2){            sum = l2->val + carry;            carry = sum / 10;            cur-> next = new ListNode(sum % 10);            cur = cur->next;            l2 = l2->next;        }        if(!l1 && !l2 && carry){            cur-> next = new ListNode(carry);            cur = cur->next;        }        return resultList;    }};

代码二

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        long long n1 = 0, n2 = 0, count = 0, sum = 0;        while(l1){            n1 = n1 + l1->val * pow(10, count);            l1 = l1->next;            count++;        }        cout <<"n1  "<< n1<< endl;        count = 0;        while(l2){            n2 = n2 + l2->val * pow(10, count);            cout << "n2  " << n2 << endl;            l2 = l2->next;            count++;        }        sum = n2 + n1;        cout << sum << endl;        struct ListNode* resultList = new ListNode(sum % 10);        struct ListNode* nowList = resultList;        sum /= 10;        while(sum > 0){            nowList-> next = new ListNode(sum % 10);            nowList = nowList->next;            sum = sum / 10;            cout << sum << endl;        }        return resultList;    }};

结果

他山之玉

C++ O(n) solutioin

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {    ListNode preHead(0), *p = &preHead;    int extra = 0;    while (l1 || l2 || extra) {        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;        extra = sum / 10;        p->next = new ListNode(sum % 10);        p = p->next;        l1 = l1 ? l1->next : l1;        l2 = l2 ? l2->next : l2;    }    return preHead.next;}

Java O(n) solution

public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode ln1 = l1, ln2 = l2, head = null, node = null;        int carry = 0, remainder = 0, sum = 0;        head = node = new ListNode(0);        while(ln1 != null || ln2 != null || carry != 0) {            sum = (ln1 != null ? ln1.val : 0) + (ln2 != null ? ln2.val : 0) + carry;            carry = sum / 10;            remainder = sum % 10;            node = node.next = new ListNode(remainder);            ln1 = (ln1 != null ? ln1.next : null);            ln2 = (ln2 != null ? ln2.next : null);        }        return head.next;    }}

python O(n) solution

class Solution:# @return a ListNodedef addTwoNumbers(self, l1, l2):    carry = 0    root = n = ListNode(0)    while l1 or l2 or carry:        v1 = v2 = 0        if l1:            v1 = l1.val            l1 = l1.next        if l2:            v2 = l2.val            l2 = l2.next        carry, val = divmod(v1+v2+carry, 10)        n.next = ListNode(val)        n = n.next    return root.next

反思

思路是相近的，不过我的还不够简介。加油，基本功语言还需要加强！！！！