338. Counting Bits
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.
Example 1:
Example 1:
Input: 2 Output: [0,1,1]
Example 2:
Input: 5
Output: [0,1,1,2,1,2]
Follow up:
- It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
- Space complexity should be O(n).
- Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
思路：题目意为给定一个非负整数，计算每个从0至num的数字二进制中1的个数，返回数组。用动态规划的思想可以举例
dp[0] = 0;
dp[1] = dp[0] + 1;
dp[2] = dp[0] + 1;
dp[3] = dp[1] +1;
dp[4] = dp[0] + 1;
dp[5] = dp[1] + 1;
dp[6] = dp[2] + 1;
dp[7] = dp[3] + 1;
dp[8] = dp[0] + 1;
等于
dp[0] = 0;
dp[1] = dp[1-1] + 1;
dp[2] = dp[2-2] + 1;
dp[3] = dp[3-2] +1;
dp[4] = dp[4-4] + 1;
dp[5] = dp[5-4] + 1;
dp[6] = dp[6-4] + 1;
dp[7] = dp[7-4] + 1;
dp[8] = dp[8-8] + 1;
能得出dp[index] = dp[index – offset] + 1;
代码如下：
public class CountingBits {
public static int[] countBits(int num) {
int[] res = new int[num + 1];
int i = 1;
int offset = 1;
while (i < num + 1) {
if (offset * 2 == i) {
offset = offset * 2;
}
res[i] = res[i - offset] + 1;
i++;
}
return res;
}
}
