README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1017.Convert%20to%20Base%20-2/README_EN.md	1697	Weekly Contest 130 Q2	Math

1017. Convert to Base -2

中文文档

Description

Given an integer n, return a binary string representing its representation in base -2.

Note that the returned string should not have leading zeros unless the string is "0".

 

Example 1:

Input: n = 2
Output: "110"
Explantion: (-2)2 + (-2)1 = 2

Example 2:

Input: n = 3
Output: "111"
Explantion: (-2)2 + (-2)1 + (-2)0 = 3

Example 3:

Input: n = 4
Output: "100"
Explantion: (-2)2 = 4

 

Constraints:

• 0 <= n <= 109

Solutions

Solution 1

Python3

class Solution:
    def baseNeg2(self, n: int) -> str:
        k = 1
        ans = []
        while n:
            if n % 2:
                ans.append('1')
                n -= k
            else:
                ans.append('0')
            n //= 2
            k *= -1
        return ''.join(ans[::-1]) or '0'

Java

class Solution {
    public String baseNeg2(int n) {
        if (n == 0) {
            return "0";
        }
        int k = 1;
        StringBuilder ans = new StringBuilder();
        while (n != 0) {
            if (n % 2 != 0) {
                ans.append(1);
                n -= k;
            } else {
                ans.append(0);
            }
            k *= -1;
            n /= 2;
        }
        return ans.reverse().toString();
    }
}

C++

class Solution {
public:
    string baseNeg2(int n) {
        if (n == 0) {
            return "0";
        }
        int k = 1;
        string ans;
        while (n) {
            if (n % 2) {
                ans.push_back('1');
                n -= k;
            } else {
                ans.push_back('0');
            }
            k *= -1;
            n /= 2;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

Go

func baseNeg2(n int) string {
	if n == 0 {
		return "0"
	}
	ans := []byte{}
	k := 1
	for n != 0 {
		if n%2 != 0 {
			ans = append(ans, '1')
			n -= k
		} else {
			ans = append(ans, '0')
		}
		k *= -1
		n /= 2
	}
	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)
}

TypeScript

function baseNeg2(n: number): string {
    if (n === 0) {
        return '0';
    }
    let k = 1;
    const ans: string[] = [];
    while (n) {
        if (n % 2) {
            ans.push('1');
            n -= k;
        } else {
            ans.push('0');
        }
        k *= -1;
        n /= 2;
    }
    return ans.reverse().join('');
}

Rust

impl Solution {
    pub fn base_neg2(n: i32) -> String {
        if n == 0 {
            return "0".to_string();
        }
        let mut k = 1;
        let mut ans = String::new();
        let mut num = n;
        while num != 0 {
            if num % 2 != 0 {
                ans.push('1');
                num -= k;
            } else {
                ans.push('0');
            }
            k *= -1;
            num /= 2;
        }
        ans.chars().rev().collect::<String>()
    }
}

C#

public class Solution {
    public string BaseNeg2(int n) {
        if (n == 0) {
            return "0";
        }
        int k = 1;
        StringBuilder ans = new StringBuilder();
        int num = n;
        while (num != 0) {
            if (num % 2 != 0) {
                ans.Append('1');
                num -= k;
            } else {
                ans.Append('0');
            }
            k *= -1;
            num /= 2;
        }
        char[] cs = ans.ToString().ToCharArray();
        Array.Reverse(cs);
        return new string(cs);
    }
}