【LeetCode】989. Add to Array-Form of Integer 数组形式的整数加法（Easy）（JAVA）
题目地址: https://leetcode.com/problems/add-to-array-form-of-integer/
题目描述：
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, then A[0] != 0
题目大意
对于非负整数 X 而言，X 的数组形式是每位数字按从左到右的顺序形成的数组。例如，如果 X = 1231，那么其数组形式为 [1,2,3,1]。
给定非负整数 X 的数组形式 A，返回整数 X+K 的数组形式。
解题方法
- 先把数组所有元素放入结果的 list 里面
- 遍历 K 知道 K 为 0 为止
class Solution {
public List<Integer> addToArrayForm(int[] A, int K) {
List<Integer> res = new ArrayList<>();
for (int i = 0; i < A.length; i++) {
res.add(A[i]);
}
int carry = 0;
for (int i = res.size() - 1; i >= 0; i--) {
carry += res.get(i);
carry += K % 10;
res.set(i, carry % 10);
K /= 10;
carry /= 10;
}
while (K > 0 || carry > 0) {
carry += K % 10;
res.add(0, carry % 10);
K /= 10;
carry /= 10;
}
return res;
}
}
执行耗时:5 ms,击败了56.68% 的Java用户
内存消耗:39.8 MB,击败了57.27% 的Java用户
