123th LeetCode Weekly Contest Add to Array-Form of Integer-优快云博客

本文详细解析了一种处理大数加法的算法，通过将整数转换为数组形式，实现了数组形式的大数相加。文章提供了具体示例，如1200+34=1234等，展示了算法的正确性和效率。同时，代码中使用了C++语言进行实现，适合对大数运算感兴趣的读者。

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For a non-negative integer X, the array-form of X is an array of its digits in left to right order. For example, if X = 1231, then the array form is [1,2,3,1].

Given the array-form A of a non-negative integer X, return the array-form of the integer X+K.

Example 1:

Input: A = [1,2,0,0], K = 34
Output: [1,2,3,4]
Explanation: 1200 + 34 = 1234

Example 2:

Input: A = [2,7,4], K = 181
Output: [4,5,5]
Explanation: 274 + 181 = 455

Example 3:

Input: A = [2,1,5], K = 806
Output: [1,0,2,1]
Explanation: 215 + 806 = 1021

Example 4:

Input: A = [9,9,9,9,9,9,9,9,9,9], K = 1
Output: [1,0,0,0,0,0,0,0,0,0,0]
Explanation: 9999999999 + 1 = 10000000000

Note：

1 <= A.length <= 10000
0 <= A[i] <= 9
0 <= K <= 10000
If A.length > 1, then A[0] != 0

这题肯定有简单解法，写大数加法只是更熟练而已

class Solution {
public:
    vector<int> addToArrayForm(vector<int>& A, int K) {
        int len = A.size();
        int num=0;
        vector<int>Ve;
        int a[11000]={0},b[11000]={0},c[11000]={0};
        for(int i=len-1;i>=0;i--){
            a[num++]=A[i];
            //cout<<A[i]<<endl;
        }
        int cnt=0;
        int i;
        while(K){
            b[cnt++]=K%10;
            K/=10;
        }
        //cout<<"B"<<endl;
        int k=max(len,cnt);
       
        for(i=0;i<k;i++)
        {
            //cout<<a[i]<<" "<<b[i]<<endl;
            c[i]=a[i]+b[i];
        }
        for(i=0; i<k; i++){
            //cout<<"A1"<<endl;
            if(c[i]>=10){
                c[i+1]+=c[i]/10;
                c[i]%=10;
            }
            //cout<<"A2"<<endl;
        }
        
        if(c[k]==0) k--;
         for(i=k;i>=0;i--){
             Ve.push_back(c[i]);
         }
             
        
        return Ve;
            
    }
};