2019-11-02 12:07:41 +08:00
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# 15. 二进制中 1 的个数
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[NowCoder](https://www.nowcoder.com/practice/8ee967e43c2c4ec193b040ea7fbb10b8?tpId=13&tqId=11164&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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输入一个整数，输出该数二进制表示中 1 的个数。
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### n&(n-1)
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该位运算去除 n 的位级表示中最低的那一位。
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```
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n : 10110100
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n-1 : 10110011
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n&(n-1) : 10110000
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```
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时间复杂度：O(M)，其中 M 表示 1 的个数。
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```java
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public int NumberOf1(int n) {
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int cnt = 0;
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while (n != 0) {
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cnt++;
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n &= (n - 1);
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}
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return cnt;
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}
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```
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### Integer.bitCount()
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```java
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public int NumberOf1(int n) {
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return Integer.bitCount(n);
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}
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```
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2019-11-02 14:39:13 +08:00
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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