LeetCode 0989. Add to Array-Form of Integer Solution in Java, C++, Python & More | Explanation + Code

Description

The array-form of an integer num is an array representing its digits in left to right order.

• For example, for num = 1321, the array form is [1,3,2,1].

Given num, the array-form of an integer, and an integer k, return the array-form of the integer num + k.

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Constraints:

• 1 <= num.length <= 104
• 0 <= num[i] <= 9
• num does not contain any leading zeros except for the zero itself.
• 1 <= k <= 104

Solutions

Solution 1: Simulation

We can start from the last digit of the array and add each digit of the array to k. Then, divide k by 10, and use the remainder as the current digit's value, with the quotient as the carry. Continue this process until the array is fully traversed and k = 0. Finally, reverse the answer array.

The time complexity is O(n), where n is the length of num. Ignoring the space consumption of the answer array, the space complexity is O(1).

PythonJavaC++GoTypeScriptRust

class Solution: def addToArrayForm(self, num: List[int], k: int) -> List[int]: ans = [] i = len(num) - 1 while i >= 0 or k: k += 0 if i < 0 else num[i] k, x = divmod(k, 10) ans.append(x) i -= 1 return ans[::-1](code-box)

class Solution { public List<Integer> addToArrayForm(int[] num, int k) { List<Integer> ans = new ArrayList<>(); for (int i = num.length - 1; i >= 0 || k > 0; --i) { k += (i >= 0 ? num[i] : 0); ans.add(k % 10); k /= 10; } Collections.reverse(ans); return ans; } }(code-box)

class Solution { public: vector<int> addToArrayForm(vector<int>& num, int k) { vector<int> ans; for (int i = num.size() - 1; i >= 0 || k > 0; --i) { k += (i >= 0 ? num[i] : 0); ans.push_back(k % 10); k /= 10; } ranges::reverse(ans); return ans; } };(code-box)

func addToArrayForm(num []int, k int) (ans []int) { for i := len(num) - 1; i >= 0 || k > 0; i-- { if i >= 0 { k += num[i] } ans = append(ans, k%10) k /= 10 } slices.Reverse(ans) return }(code-box)

function addToArrayForm(num: number[], k: number): number[] { const ans: number[] = []; for (let i = num.length - 1; i >= 0 || k > 0; --i) { k += i >= 0 ? num[i] : 0; ans.push(k % 10); k = Math.floor(k / 10); } return ans.reverse(); }(code-box)

impl Solution { pub fn add_to_array_form(num: Vec<i32>, k: i32) -> Vec<i32> { let mut ans = Vec::new(); let mut k = k; let mut i = num.len() as i32 - 1; while i >= 0 || k > 0 { if i >= 0 { k += num[i as usize]; } ans.push(k % 10); k /= 10; i -= 1; } ans.reverse(); ans } }(code-box)