[LeetCode][338]counting-bits
Content
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_popcountin C++)?
Related Topics
Solution
1. 位运算
Java
class Solution {
public int[] countBits(int n) {
// 0 <= n <= 10⁵
int[] ans = new int[n + 1];
for (int i = 0; i <= n; i++) {
int j = i;
while (j > 0) {
ans[i] += j & 1;
j >>= 1;
}
}
return ans;
}
}
2. 动态规划
Java
class Solution {
public int[] countBits(int n) {
// 0 <= n <= 10⁵
int[] ans = new int[n + 1];
for (int i = 0; i <= n; i++) {
ans[i] = (i & 1) + ans[i >> 1];
}
return ans;
}
}