338.Counting Bits
338. Counting Bits
- Given an integer
n
, return an arrayans
of lengthn + 1
such that for eachi
(0 <= i <= n
),ans[i]
is the number of1
’s in the binary representation ofi
.
Example 1
Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10
Example 2
Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101
Method 1
【O(n) time | O(n) space】
package Leetcode.BitManipulation;
import java.util.Arrays;
/**
* @author zhengstars
* @date 2024/06/09
*/
public class CountingBits {
public static int[] countBits(int n) {
// Initialize the result array with length n + 1.
int[] ans = new int[n + 1];
// Loop through the numbers from 1 to n.
for (int i = 1; i <= n; i++) {
// The number of 1's in the binary representation of i is equal to:
// the number of 1's in i / 2 (i.e., i >> 1) plus
// 1 if the least significant bit of i is 1 (i.e., i & 1).
// This is based on the dynamic programming principle where the solution
// for a larger problem is built from the solutions of smaller subproblems.
ans[i] = ans[i >> 1] + (i & 1);
}
// Return the result array.
return ans;
}
public static void main(String[] args) {
// Test case 1
int[] result1 = countBits(2);
System.out.println(Arrays.toString(result1)); // Expected: [0, 1, 1]
// Test case 2
int[] result2 = countBits(5);
System.out.println(Arrays.toString(result2)); // Expected: [0, 1, 1, 2, 1, 2]
}
}
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