LeetCode 2595. Number of Even and Odd Bits
原题链接在这里:https://leetcode.com/problems/number-of-even-and-odd-bits/description/
题目:
You are given a positive integer n
.
Let even
denote the number of even indices in the binary representation of n
(0-indexed) with value 1
.
Let odd
denote the number of odd indices in the binary representation of n
(0-indexed) with value 1
.
Return an integer array answer
where answer = [even, odd]
.
Example 1:
Input: n = 17 Output: [2,0] Explanation: The binary representation of 17 is 10001. It contains 1 on the 0th and 4th indices. There are 2 even and 0 odd indices.
Example 2:
Input: n = 2 Output: [0,1] Explanation: The binary representation of 2 is 10. It contains 1 on the 1st index. There are 0 even and 1 odd indices.
Constraints:
1 <= n <= 1000
题解:
Have even and odd count, when n & cur != 0, accumulate the corresponding count.
Time Complexity: O(1). Since it is interger, it can't be more than 32 bits.
Space: O(1).
AC Java:
1 class Solution { 2 public int[] evenOddBit(int n) { 3 int even = 0; 4 int odd = 0; 5 int cur = 1; 6 boolean isEven = true; 7 while(cur <= n){ 8 if((n & cur) != 0){ 9 if(isEven){ 10 even++; 11 }else{ 12 odd++; 13 } 14 } 15 16 cur = cur << 1; 17 isEven = !isEven; 18 } 19 20 return new int[]{even, odd}; 21 } 22 }