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
Runtime:
136 ms, faster than 100.00% of C++ online submissions for Add to Array-Form of Integer.
Memory Usage:
11.9 MB, less than 100.00% of C++ online submissions for Add to Array-Form of Integer.
class Solution { public: vector<int> addToArrayForm(vector<int>& A, int K) { vector<int> ka = itoarray(K); for(int a : ka) cout << a << endl; reverse(A.begin(), A.end()); vector<int> ret; int endi = max(A.size(), ka.size()); int add1, add2, sleft, sj; sj = 0; for(int i=0; i<endi; i++) { add1 = add2 = sleft = 0; if(i < A.size()) add1 = A[i]; if(i < ka.size()) add2 = ka[i]; sleft = sj + add1 + add2; if(sleft >= 10) {sj = 1; sleft %= 10;} else sj = 0; ret.push_back(sleft); } if(sj == 1) ret.push_back(1); reverse(ret.begin(), ret.end()); return ret; } vector<int> itoarray(int K) { vector<int> a; while(K != 0) { a.push_back(K % 10); K /= 10; } return a; } };
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