338. Counting Bits
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 < = 1 0 5 0 <= n <= 10^5 0<=n<=105
From: LeetCode
Link: 338. Counting Bits
Solution:
Ideas：
This function first allocates memory for an array of size n + 1 to store the counts. It initializes the first element of the array with 0, as the binary representation of 0 contains 0 ones. Then, for each number from 1 to n, it calculates the number of 1’s based on the observation mentioned above. Specifically, it uses right shift (i >> 1) to divide the number by 2 and uses bitwise AND with 1 (i & 1) to determine if the number is odd (in which case one more 1 must be added). Finally, it returns the populated array and sets the return size to n + 1.
Caode:
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* countBits(int n, int* returnSize) {
*returnSize = n + 1; // Set the return size.
int* ans = (int*)malloc((*returnSize) * sizeof(int)); // Allocate memory for the answer array.
ans[0] = 0; // The number of 1's in 0 is 0.
for (int i = 1; i <= n; i++) {
// If i is even, then i and i/2 have the same number of 1's in their binary representation.
// If i is odd, then i has one more 1 than i - 1 in its binary representation.
ans[i] = ans[i >> 1] + (i & 1);
}
return ans;
}
