Counting Bits

Description

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.

Solution(javascript)

/*
 * @lc app=leetcode id=338 lang=javascript
 *
 * [338] Counting Bits
 */

// @lc code=start
// /** 1: 逐个计算
//  * @param {number} num
//  * @return {number[]}
//  */
// const countBits = function (num) {
//   const count = (n) => {
//     let result = 0
//     while (n) {
//       n &= (n - 1)
//       result += 1
//     }
//     return result
//   }
//   const arr = []
//   for (let i = 0; i <= num; i++) {
//     arr[i] = count(i)
//   }
//   return arr
// }

// /** 2.1: DP
//  * @param {number} num
//  * @return {number[]}
//  */
// const countBits = function (num) {
//   const dp = [0]
//   for (let i = 1; i <= num; i++) {
//     dp[i] = dp[i & (i - 1)] + 1
//   }
//   return dp
// }

/** 2.2: DP
 * @param {number} num
 * @return {number[]}
 */
const countBits = function (num) {
  const dp = [0]
  for (let i = 1; i <= num; i++) {
    if ((i & 1) === 0) {
      dp[i] = dp[i >> 1]
    } else {
      dp[i] = dp[i - 1] + 1
    }
  }
  return dp
}
// @lc code=end