338. Counting Bits #

题目 #

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:

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.

题目大意 #

给定一个非负整数 num。对于 0 ≤ i ≤ num 范围中的每个数字 i ，计算其二进制数中的 1 的数目并将它们作为数组返回。

解题思路 #

• 给出一个数，要求计算出 0 ≤ i ≤ num 中每个数的二进制位 1 的个数。

• 这一题就是利用二进制位运算的经典题。

    X&1==1or==0，可以用 X&1 判断奇偶性，X&1>0 即奇数。
    X = X & (X-1) 清零最低位的1
    X & -X => 得到最低位的1 
    X&~X=>0
  

代码 #

package leetcode

func countBits(num int) []int {
	bits := make([]int, num+1)
	for i := 1; i <= num; i++ {
		bits[i] += bits[i&(i-1)] + 1
	}
	return bits
}