Problem
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:
0 <= n <= 105
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 timeO(n)
and possibly in a single pass?Can you do it without using any built-in function (i.e., like
__builtin_popcount
in C++)?
Solution
/**
* @param {number} n
* @return {number[]}
*/
var countBits = function(n) {
var res = Array(n + 1);
var num = 2;
var nextNum = 4;
for (var i = 0; i <= n; i++) {
if (i === 0) {
res[i] = 0;
} else if (i === 1) {
res[i] = 1;
} else {
if (i === nextNum) {
num = nextNum;
nextNum *= 2;
}
res[i] = 1 + res[i - num];
}
}
return res;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).