README_EN.md

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

Python3

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])

Java

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();
    }
}

C++

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;
    }
};

Go

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)
}

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