# Count 1 in Binary ### Source - lintcode: [(365) Count 1 in Binary](http://www.lintcode.com/en/problem/count-1-in-binary/) ~~~ Count how many 1 in binary representation of a 32-bit integer. Example Given 32, return 1 Given 5, return 2 Given 1023, return 9 Challenge If the integer is n bits with m 1 bits. Can you do it in O(m) time? ~~~ ### 題解 題 [O1 Check Power of 2](http://algorithm.yuanbin.me/zh-cn/math_and_bit_manipulation/o1_check_power_of_2.html) 的進階版，`x & (x - 1)` 的含義為去掉二進制數中1的最后一位，無論 x 是正數還是負數都成立。 ### Java ~~~ public class Solution { /** * @param num: an integer * @return: an integer, the number of ones in num */ public int countOnes(int num) { int count = 0; while (num != 0) { num = num & (num - 1); count++; } return count; } } ~~~ ### 源碼分析 累加計數器即可。 ### 復雜度分析 這種算法依賴于數中1的個數，時間復雜度為 O(m)O(m)O(m). 空間復雜度 O(1)O(1)O(1). ### Reference - [Number of 1 bits | LeetCode](http://articles.leetcode.com/2010/09/number-of-1-bits.html) - 評論中有關于不同算法性能的討論