LeetCode-Add to Array-Form of Integer

来源：恒创科技编辑：恒创科技编辑部

2024-02-03 07:57:59

Description：
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 <= 100000 <= A[i] <= 90 <= K <= 10000If A.length > 1, then A[0] != 0

题意：给定一个数组A，里面每个元素的范围在[0, 9]；和一个整数K，对数组A执行累加K的操作，从最低为开始执行加法，并保证相加后每一位的范围仍在[0, 9]；

解法：其实就是让我们实现一个大数加法，只不过现在一个以数组的形式给出，一个直接给出整数；定义最终的返回结果为res；
第一步：同时从A和K的最低位开始遍历，存储相加后的结果到res中(注意：需要添加在首部，因此我们利用LinkedList实现)；
第二步：完成第一步后会有两种可能的情况：
1）如果数组A遍历完后，K依然大于0，此时需要将余下的K添加到res中
2）如果此时进位不为0，那么也需要添加到res中

Java

class Solution {
    public List<Integer> addToArrayForm(int[] A, int K) {
        LinkedList<Integer> res = new LinkedList<>();
        int carry = 0;
        for (int i = A.length - 1; i >= 0; i--) {
            int x = A[i] + carry + K % 10;
            carry = x / 10;
            x %= 10;
            K /= 10;
            res.addFirst(x);
        }
        while (K > 0) {
            int x = carry + K % 10;
            carry = x / 10;
            x %= 10;
            K /= 10;
            res.addFirst(x);
        }
        if (carry > 0) {
            res.addFirst(carry);
        }
        
        return res;
    }
}