Description
Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.
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
Constraints:
Follow up:
- It is very easy to come up with a solution with a runtime of
O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
- Can you do it without using any built-in function (i.e., like
__builtin_popcount in C++)?
Solutions
Solution 1
PythonJavaC++GoTypeScript
class Solution:
def countBits(self, n: int) -> List[int]:
return [i.bit_count() for i in range(n + 1)](code-box)
class Solution {
public int[] countBits(int n) {
int[] ans = new int[n + 1];
for (int i = 0; i <= n; ++i) {
ans[i] = Integer.bitCount(i);
}
return ans;
}
}(code-box)
class Solution {
public:
vector<int> countBits(int n) {
vector<int> ans(n + 1);
for (int i = 0; i <= n; ++i) {
ans[i] = __builtin_popcount(i);
}
return ans;
}
};(code-box)
func countBits(n int) []int {
ans := make([]int, n+1)
for i := 0; i <= n; i++ {
ans[i] = bits.OnesCount(uint(i))
}
return ans
}(code-box)
function countBits(n: number): number[] {
const ans: number[] = Array(n + 1).fill(0);
for (let i = 0; i <= n; ++i) {
ans[i] = bitCount(i);
}
return ans;
}
function bitCount(n: number): number {
let count = 0;
while (n) {
n &= n - 1;
++count;
}
return count;
}(code-box)
Solution 2
PythonJavaC++GoTypeScript
class Solution:
def countBits(self, n: int) -> List[int]:
ans = [0] * (n + 1)
for i in range(1, n + 1):
ans[i] = ans[i & (i - 1)] + 1
return ans(code-box)
class Solution {
public int[] countBits(int n) {
int[] ans = new int[n + 1];
for (int i = 1; i <= n; ++i) {
ans[i] = ans[i & (i - 1)] + 1;
}
return ans;
}
}(code-box)
class Solution {
public:
vector<int> countBits(int n) {
vector<int> ans(n + 1);
for (int i = 1; i <= n; ++i) {
ans[i] = ans[i & (i - 1)] + 1;
}
return ans;
}
};(code-box)
func countBits(n int) []int {
ans := make([]int, n+1)
for i := 1; i <= n; i++ {
ans[i] = ans[i&(i-1)] + 1
}
return ans
}(code-box)
function countBits(n: number): number[] {
const ans: number[] = Array(n + 1).fill(0);
for (let i = 1; i <= n; ++i) {
ans[i] = ans[i & (i - 1)] + 1;
}
return ans;
}(code-box)
