对于非负整数 X 而言,X 的数组形式是每位数字按从左到右的顺序形成的数组。例如,如果 X = 1231,那么其数组形式为 [1,2,3,1]。
给定非负整数 X 的数组形式 A,返回整数 X+K 的数组形式。
示例 1:
输入:A = [1,2,0,0], K = 34
输出:[1,2,3,4]
解释:1200 + 34 = 1234
示例 2:
输入:A = [2,7,4], K = 181
输出:[4,5,5]
解释:274 + 181 = 455
示例 3:
输入:A = [2,1,5], K = 806
输出:[1,0,2,1]
解释:215 + 806 = 1021
示例 4:
输入:A = [9,9,9,9,9,9,9,9,9,9], K = 1
输出:[1,0,0,0,0,0,0,0,0,0,0]
解释:9999999999 + 1 = 10000000000
提示:
1 <= A.length <= 10000
0 <= A[i] <= 9
0 <= K <= 10000
如果 A.length > 1,那么 A[0] != 0
接着用大整数类,只是增加了两个构造函数,爽!
#define MAX 100000
class MyInt{
private:
int _n;
int *_arr;
public:
MyInt();
MyInt(int k);
MyInt(vector<int> v);
MyInt(string s); // 支持从字符串创建
friend MyInt operator+(MyInt const &mi1, MyInt const &mi2);
string toString();
vector<int> toVector();
friend void print(MyInt mi);
~MyInt();
};
MyInt::MyInt(){
_n=0;
_arr = new int[MAX];
}
MyInt::MyInt(int k){
_n = 0;
_arr = new int[MAX];
if(k == 0){
_n = 1;
_arr[0] = 0;
}
while(k!=0){
_arr[_n] = k%10;
k /= 10;
_n++;
}
}
MyInt::MyInt(vector<int> v){
_n = v.size();
_arr = new int[MAX];
for(int i=0; i<_n; i++)
_arr[i] = v[_n-1-i];
}
// 倒着存
MyInt::MyInt(string s){
_n = s.length();
_arr = new int[MAX];
for(int i=0; i<_n; i++)
_arr[i] = s[_n-1-i]-'0';
}
MyInt operator+(MyInt const &mi1, MyInt const &mi2){
MyInt mi3;
int i1=0, i2=0, i3=0;
bool c = false; // 进位标志
while(i1<mi1._n || i2<mi2._n){
int k1 = i1<mi1._n?mi1._arr[i1]:0;
int k2 = i2<mi2._n?mi2._arr[i2]:0;
int t = k1+k2+c;
if(t >= 10){
t %= 10;
c = true;
}
else c = false;
mi3._arr[i3] = t;
i1++; i2++; i3++;
}
mi3._arr[i3]=c;
mi3._n = i3+c;
return mi3;
}
string MyInt::toString(){
string s;
for(int i=0; i<_n; i++)
s += _arr[_n-1-i]+'0';
return s;
}
vector<int> MyInt::toVector(){
vector<int> v;
for(int i=0; i<_n; i++)
v.push_back(_arr[_n-1-i]);
return v;
}
void print(MyInt mi){
for(int i=0; i<mi._n; i++)
cout<<mi._arr[mi._n-1-i];
}
MyInt::~MyInt(){
delete[] _arr;
}
class Solution {
public:
vector<int> addToArrayForm(vector<int>& A, int K) {
MyInt mi1(A);
MyInt mi2(K);
MyInt mi3 = mi1+mi2;
return mi3.toVector();
}
};