題目

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. 1 <= A.length <= 10000
  2. 0 <= A[i] <= 9
  3. 0 <= K <= 10000
  4. If A.length > 1, then A[0] != 0

題目大意

對於非負整數 X 而言，X 的數組形式是每位數字按從左到右的順序形成的數組。例如，如果 X = 1231，那麼其數組形式為 [1,2,3,1]。給定非負整數 X 的數組形式 A，返回整數 X+K 的數組形式。

解題思路

  • 簡單題，計算 2 個非負整數的和。累加過程中不斷的進位，最終輸出到數組中記得需要逆序，即數字的高位排在數組下標較小的位置。

參考代碼

package leetcode

func addToArrayForm(A []int, K int) []int {
	res := []int{}
	for i := len(A) - 1; i >= 0; i-- {
		sum := A[i] + K%10
		K /= 10
		if sum >= 10 {
			K++
			sum -= 10
		}
		res = append(res, sum)
	}
	for ; K > 0; K /= 10 {
		res = append(res, K%10)
	}
	reverse(res)
	return res
}

func reverse(A []int) {
	for i, n := 0, len(A); i < n/2; i++ {
		A[i], A[n-1-i] = A[n-1-i], A[i]
	}
}