Description#
Given an array of positive integers nums, return an array answer that consists of the digits of each integer in nums after separating them in the same order they appear in nums.
To separate the digits of an integer is to get all the digits it has in the same order.
- For example, for the integer
10921, the separation of its digits is [1,0,9,2,1].
Example 1:
Input: nums = [13,25,83,77]
Output: [1,3,2,5,8,3,7,7]
Explanation:
- The separation of 13 is [1,3].
- The separation of 25 is [2,5].
- The separation of 83 is [8,3].
- The separation of 77 is [7,7].
answer = [1,3,2,5,8,3,7,7]. Note that answer contains the separations in the same order.
Example 2:
Input: nums = [7,1,3,9]
Output: [7,1,3,9]
Explanation: The separation of each integer in nums is itself.
answer = [7,1,3,9].
Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 105
Solutions#
Solution 1: Simulation#
Split each number in the array into digits, then put the split numbers into the answer array in order.
The time complexity is $O(n \times \log_{10} M)$, and the space complexity is $O(n \times \log_{10} M)$. Where $n$ is the length of the array $nums$, and $M$ is the maximum value in the array $nums$.
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| class Solution:
def separateDigits(self, nums: List[int]) -> List[int]:
ans = []
for x in nums:
t = []
while x:
t.append(x % 10)
x //= 10
ans.extend(t[::-1])
return ans
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| class Solution {
public int[] separateDigits(int[] nums) {
List<Integer> res = new ArrayList<>();
for (int x : nums) {
List<Integer> t = new ArrayList<>();
for (; x > 0; x /= 10) {
t.add(x % 10);
}
Collections.reverse(t);
res.addAll(t);
}
int[] ans = new int[res.size()];
for (int i = 0; i < ans.length; ++i) {
ans[i] = res.get(i);
}
return ans;
}
}
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| class Solution {
public:
vector<int> separateDigits(vector<int>& nums) {
vector<int> ans;
for (int x : nums) {
vector<int> t;
for (; x; x /= 10) {
t.push_back(x % 10);
}
while (t.size()) {
ans.push_back(t.back());
t.pop_back();
}
}
return ans;
}
};
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| func separateDigits(nums []int) (ans []int) {
for _, x := range nums {
t := []int{}
for ; x > 0; x /= 10 {
t = append(t, x%10)
}
for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
t[i], t[j] = t[j], t[i]
}
ans = append(ans, t...)
}
return
}
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| function separateDigits(nums: number[]): number[] {
const ans: number[] = [];
for (let num of nums) {
const t: number[] = [];
while (num) {
t.push(num % 10);
num = Math.floor(num / 10);
}
ans.push(...t.reverse());
}
return ans;
}
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| impl Solution {
pub fn separate_digits(nums: Vec<i32>) -> Vec<i32> {
let mut ans = Vec::new();
for &num in nums.iter() {
let mut num = num;
let mut t = Vec::new();
while num != 0 {
t.push(num % 10);
num /= 10;
}
t.into_iter()
.rev()
.for_each(|v| ans.push(v));
}
ans
}
}
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| /**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* separateDigits(int* nums, int numsSize, int* returnSize) {
int n = 0;
for (int i = 0; i < numsSize; i++) {
int t = nums[i];
while (t != 0) {
t /= 10;
n++;
}
}
int* ans = malloc(sizeof(int) * n);
for (int i = numsSize - 1, j = n - 1; i >= 0; i--) {
int t = nums[i];
while (t != 0) {
ans[j--] = t % 10;
t /= 10;
}
}
*returnSize = n;
return ans;
}
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Solution 2#
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| impl Solution {
pub fn separate_digits(nums: Vec<i32>) -> Vec<i32> {
let mut ans = vec![];
for n in nums {
let mut t = vec![];
let mut x = n;
while x != 0 {
t.push(x % 10);
x /= 10;
}
for i in (0..t.len()).rev() {
ans.push(t[i]);
}
}
ans
}
}
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