标签：res nbsp || ret 直接 lis ide lan RKE
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 <= 10000A.length > 1, then A[0] != 0Idea 1: 大数的相加
Time complexity: O(max(n, log(K))), n = A.length
Space complexity: O(max(n, log(K)))
1 class Solution { 2 public List<Integer> addToArrayForm(int[] A, int K) { 3 List<Integer> result = new ArrayList<>(); 4 int carry = 0; 5 6 for(int left = A.length-1, curr = K; left >= 0 || curr != 0 || carry != 0; --left) { 7 int val1 = left >= 0? A[left] : 0; 8 int val2 = curr%10; 9 int sum = val1 + val2 + carry; 10 result.add(sum%10); 11 carry = sum/10; 12 curr = curr/10; 13 } 14 15 Collections.reverse(result); 16 return result; 17 } 18 }
Idea 1.a 网上的解法，代码更简洁，少用carry这个变量，直接用K加数组里的每个数字
Note. 如果K是Integer.MAX_VALUE, 这样加会overflow
A= {1, 2, 9, 9}, K = 99
1 class Solution { 2 public List<Integer> addToArrayForm(int[] A, int K) { 3 List<Integer> result = new ArrayList<>(); 4 int carry = 0; 5 6 for(int left = A.length-1, curr = K; left >= 0 || curr != 0 ; --left) { 7 if(left >= 0) { 8 curr += A[left]; 9 } 10 11 result.add(curr%10); 12 curr = curr/10; 13 } 14 15 Collections.reverse(result); 16 return result; 17 } 18 }
Add to Array-Form of Integer LT989
标签：res nbsp || ret 直接 lis ide lan RKE
原文地址：https://www.cnblogs.com/taste-it-own-it-love-it/p/10657888.html
