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:
O(n log n). Can you do it in linear time O(n) and possibly in a single pass?__builtin_popcount in C++)?class Solution {
public int[] countBits(int n) {
if (n == 0)
return new int[]{0};
int[] dp = new int[n + 1];
dp[1] = 1;
for (int i = 1, ceil = 1; i <= n; i++) {
if (i == ceil * 2) {
ceil *= 2;
dp[i] = 1;
} else
dp[i] = dp[ceil] + dp[i - ceil];
}
return dp;
}
}