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 = 1234Example 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 <= 100000 <= A[i] <= 90 <= K <= 10000- If
A.length > 1, thenA[0] != 0
整数相加。这题跟[LeetCode] 66. Plus One非常类似。题意是用一个数组A（表示了一个非负整数）和一个整数K的相加，求和并返回一个数组。我的思路是从数组的最右边的那一位开始每一位和K相加，把相加结果%10（其实也就是相加结果的个位数）放进res，最后再将res反转。注意K有可能有多个digit，需要额外判断是不是K == 0了再跳出while循环（Java第8行）。
时间O(n)
空间O(1) - 没有创造额外空间
Java实现
1 class Solution { 2 public List<Integer> addToArrayForm(int[] A, int K) { 3 int len = A.length; 4 int cur = K; 5 List<Integer> res = new ArrayList<>(); 6 7 int i = len - 1; 8 while (i >= 0 || cur > 0) { 9 if (i >= 0) { 10 cur += A[i]; 11 } 12 res.add(cur % 10); 13 cur /= 10; 14 i--; 15 } 16 Collections.reverse(res); 17 return res; 18 } 19 }
JavaScript实现
1 /** 2 * @param {number[]} A 3 * @param {number} K 4 * @return {number[]} 5 */ 6 var addToArrayForm = function(A, K) { 7 const len = A.length; 8 let cur = K; 9 let res = []; 10 let i = len - 1; 11 while (i >= 0 || cur > 0) { 12 if (i >= 0) { 13 cur += A[i]; 14 } 15 res.push(cur % 10); 16 cur = parseInt(cur / 10); 17 i--; 18 } 19 return res.reverse(); 20 };