# 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? ~~~ ### 题解 题 [O1 Check Power of 2](http://algorithm.yuanbin.me/zh-cn/math_and_bit_manipulation/o1_check_power_of_2.html) 的进阶版，`x & (x - 1)` 的含义为去掉二进制数中1的最后一位，无论 x 是正数还是负数都成立。 ### 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; } } ~~~ ### 源码分析 累加计数器即可。 ### 复杂度分析 这种算法依赖于数中1的个数，时间复杂度为 O(m)O(m)O(m). 空间复杂度 O(1)O(1)O(1). ### Reference - [Number of 1 bits | LeetCode](http://articles.leetcode.com/2010/09/number-of-1-bits.html) - 评论中有关于不同算法性能的讨论