leetcode@ [2/43] Add Two Numbers / Multiply Strings(大整数运算)

https://leetcode.com/problems/multiply-strings/

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

class Solution {
public:
    string multiply(string num1, string num2) {
        vector<int> a, b;

        for(int i=num1.length()-;i>=;--i) a.push_back(num1[i] - '');
        for(int i=num2.length()-;i>=;--i) b.push_back(num2[i] - '');

        vector<int> res(a.size()+b.size());
        for(int i=;i<res.size();++i) res[i] = ;

        for(int i=;i<a.size();++i) {
            for(int j=;j<b.size();++j) {
                res[i+j] += a[i] * b[j];
            }
        }

        for(int i=;i<res.size();++i) {
            if(res[i]>=) {
                res[i+] += res[i]/;
                res[i] = res[i]%;
            }
        }
        //for(int i=0;i<res.size();++i) cout<<res[i]<<" ";
        //cout<<endl;
        while(res[res.size()-] == ) {
            if(res.size() == ) break;
            res.pop_back();
        }

        string sres = "";
        for(int i=res.size()-;i>=;--i) sres += res[i] + '';
        return sres;
    }
};

https://leetcode.com/problems/add-two-numbers/

You are given two linked lists representing two non-negative numbers. 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.

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

/**
 * 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) {
        vector<int> na; na.clear();
        vector<int> nb; nb.clear();

        while(l1) {
            na.push_back(l1->val);
            l1 = l1->next;
        }

        while(l2) {
            nb.push_back(l2->val);
            l2 = l2->next;
        }

        vector<int> sum(max(na.size(), nb.size()) + );
        for(int i=;i<sum.size();++i) sum[i] = ;

        int i=;
        for(;i<min(na.size(), nb.size());++i) sum[i] = na[i] + nb[i];
        for(;i<max(na.size(), nb.size());++i) sum[i] = max(na.size(), nb.size())==na.size() ? na[i]: nb[i];

        for(int i=;i<sum.size()-;++i) {
            if(sum[i] >= ) {
                sum[i+] += sum[i] / ;
                sum[i] = sum[i] % ;
            }
        }

        if(sum[sum.size()-] == ) sum.pop_back();

        ListNode *head = new ListNode(-);
        ListNode *r = head, *s;
        for(int i=;i<sum.size();++i) {
            s = new ListNode(-);
            s->val = sum[i];
            r->next = s;
            r = s;
        }r->next = NULL;

        head = head->next;

        return head;
    }
};