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1017. Convert to Base -2

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

• Java
• C++
• Python
• Go
• TypeScript

• 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();
      }
  }
  

• 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;
      }
  };
  

• 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'
  
  

• 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)
  }
  

• 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('');
  }
  
  

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