Counting Bits
Difficulty: EasyTopic: Dynamic ProgrammingBit ManipulationLeetcode:338.Counting BitsGiven 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_popcountin C++)?
Solution
write solution in here
/**
* @param {number} n
* @return {number[]}
*/
var countBits = function(n) {
const ans = new Array(n + 1).fill(0);
for (let i = 1; i <= n; i++) {
ans[i] = ans[i >> 1] + (i & 1);
}
return ans;
};