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191.number-of-1-bits.js
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/*
* @lc app=leetcode id=191 lang=javascript
*
* [191] Number of 1 Bits
*
* https://leetcode.com/problems/number-of-1-bits/description/
*
* algorithms
* Easy (48.30%)
* Likes: 825
* Dislikes: 512
* Total Accepted: 353K
* Total Submissions: 721.3K
* Testcase Example: '00000000000000000000000000001011'
*
* Write a function that takes an unsigned integer and return the number of '1'
* bits it has (also known as the Hamming weight).
*
*
*
* Example 1:
*
*
* Input: 00000000000000000000000000001011
* Output: 3
* Explanation: The input binary string 00000000000000000000000000001011 has a
* total of three '1' bits.
*
*
* Example 2:
*
*
* Input: 00000000000000000000000010000000
* Output: 1
* Explanation: The input binary string 00000000000000000000000010000000 has a
* total of one '1' bit.
*
*
* Example 3:
*
*
* Input: 11111111111111111111111111111101
* Output: 31
* Explanation: The input binary string 11111111111111111111111111111101 has a
* total of thirty one '1' bits.
*
*
*
* Note:
*
*
* Note that in some languages such as Java, there is no unsigned integer type.
* In this case, the input will be given as signed integer type and should not
* affect your implementation, as the internal binary representation of the
* integer is the same whether it is signed or unsigned.
* In Java, the compiler represents the signed integers using 2's complement
* notation. Therefore, in Example 3 above the input represents the signed
* integer -3.
*
*
*
*
* Follow up:
*
* If this function is called many times, how would you optimize it?
*
*/
// @lc code=start
/**
* @param {number} n - a positive integer
* @return {number}
*/
//simplest
var hammingWeight = function (n) {
console.log({ n, binary: n.toString(2) });
let weight = 0;
let binaryN = n.toString(2);
for (let i = 0; i < binaryN.length; i++) {
// console.log({ weight, binaryChar: binaryN.charAt(i) });
if (binaryN.charAt(i) == "1") {
weight++;
}
}
return weight;
};
// not yet know
// var hammingWeight = function (n) {
// let sum = 0;
// while (n != 0) {
// sum += n & 1;
// n = n >>> 1;
// }
// return sum;
// };
// @lc code=end