原题链接在这里:https://leetcode.com/problems/convert-to-base-2/description/
题目:
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
题解:
If it is to convert to base 2, we know append (n & 1) and then n >> 1.
For base -2, we just need to use n = -(n >> 1).
Here is the reason,
n = (a3) * (-2)^3 + (a2) * (-2) ^ 2 + (a1) * (-2)^1+ (a0) * (-2)^0
n = (-a3)*2^3 + (a2)*2^2+ (-a1)*2^1 + (a0)*2^0
By n & 1, we get a0. by n >> 1, we move to (-a1), by n = -n, we get a1.
n= (-a3)*2^3+(a2)*2^2+(-a1)*2^1+(a0)*2^0 we get a0
-n= (a3)*2^2+ (-a2)*2^1+(a1)*2^0 we get a1
n= (-a3)*2^1+ (a2)*2^0 we get a2
-n= (a3)*2^0 we get a3
Time Complexity: O(logn).
Space: O(logn).
AC Java:
1 class Solution { 2 public String baseNeg2(int n) { 3 if(n == 0){ 4 return "0"; 5 } 6 7 StringBuilder sb = new StringBuilder(); 8 while(n != 0){ 9 sb.append(n & 1); 10 n = -(n >> 1); 11 } 12 13 return sb.reverse().toString(); 14 } 15 }
标签:Convert,get,a1,a0,a3,Base,base,1017,a2 From: https://www.cnblogs.com/Dylan-Java-NYC/p/18340293