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
, thenA[0] != 0
Related Topics:
Array
Similar Questions:
// OJ: https://leetcode.com/problems/add-to-array-form-of-integer/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> addToArrayForm(vector<int>& A, int K) {
int N = A.size(), carry = 0;
vector<int> ans;
for (int i = N - 1; i >= 0 || K || carry;) {
if (K) {
carry += K % 10;
K /= 10;
}
if (i >= 0) carry += A[i--];
ans.push_back(carry % 10);
carry /= 10;
}
reverse(begin(ans), end(ans));
return ans;
}
};
Or use K
as carry
.
// OJ: https://leetcode.com/problems/add-to-array-form-of-integer/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> addToArrayForm(vector<int>& A, int K) {
int N = A.size();
vector<int> ans;
for (int i = N - 1; i >= 0 || K;) {
if (i >= 0) K += A[i--];
ans.push_back(K % 10);
K /= 10;
}
reverse(begin(ans), end(ans));
return ans;
}
};