Question

2553. Separate the Digits in an Array

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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 <= 1000
• 1 <= 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 $nums$.

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

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;
  }
}

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;
  }
};

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
}

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;
}

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
  }
}

/**
 * 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;
}

Solution 2

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
  }
}