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1556. Thousand Separator
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
-
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; } }; -
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]) -
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) }