Question

1556. Thousand Separator

中文文档

Description

Given an integer n, add a dot (".") as the thousands separator and return it in string format.

 

Example 1:

Input: n = 987
Output: "987"

Example 2:

Input: n = 1234
Output: "1.234"

 

Constraints:

• 0 <= n <= 231 - 1

Solutions

Solution 1

class Solution:
  def thousandSeparator(self, n: int) -> str:
    cnt = 0
    ans = []
    while 1:
      n, v = divmod(n, 10)
      ans.append(str(v))
      cnt += 1
      if n == 0:
        break
      if cnt == 3:
        ans.append('.')
        cnt = 0
    return ''.join(ans[::-1])

class Solution {
  public String thousandSeparator(int n) {
    int cnt = 0;
    StringBuilder ans = new StringBuilder();
    while (true) {
      int v = n % 10;
      n /= 10;
      ans.append(v);
      ++cnt;
      if (n == 0) {
        break;
      }
      if (cnt == 3) {
        ans.append('.');
        cnt = 0;
      }
    }
    return ans.reverse().toString();
  }
}

class Solution {
public:
  string thousandSeparator(int n) {
    int cnt = 0;
    string ans;
    while (1) {
      int v = n % 10;
      n /= 10;
      ans += to_string(v);
      if (n == 0) break;
      if (++cnt == 3) {
        ans += '.';
        cnt = 0;
      }
    }
    reverse(ans.begin(), ans.end());
    return ans;
  }
};

func thousandSeparator(n int) string {
  cnt := 0
  ans := []byte{}
  for {
    v := n % 10
    n /= 10
    ans = append(ans, byte('0'+v))
    if n == 0 {
      break
    }
    cnt++
    if cnt == 3 {
      ans = append(ans, '.')
      cnt = 0
    }
  }
  for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
    ans[i], ans[j] = ans[j], ans[i]
  }
  return string(ans)
}