From 2e03e232061708e3616b51d84e4643c120c1cd92 Mon Sep 17 00:00:00 2001 From: CyC2018 Date: Thu, 5 Nov 2020 00:45:34 +0800 Subject: [PATCH] auto commit --- docs/notes/15. 二进制中 1 的个数.md | 25 ++++++------------------- notes/15. 二进制中 1 的个数.md | 25 ++++++------------------- notes/pics/image-20201105004127554.png | Bin 0 -> 14598 bytes 3 files changed, 12 insertions(+), 38 deletions(-) create mode 100644 notes/pics/image-20201105004127554.png diff --git a/docs/notes/15. 二进制中 1 的个数.md b/docs/notes/15. 二进制中 1 的个数.md index 5afa17a2..d3532e14 100644 --- a/docs/notes/15. 二进制中 1 的个数.md +++ b/docs/notes/15. 二进制中 1 的个数.md @@ -1,22 +1,18 @@ # 15. 二进制中 1 的个数 -[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) +## 题目链接 + +[牛客网](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) ## 题目描述 输入一个整数，输出该数二进制表示中 1 的个数。 -### n&(n-1) +### 解题思路 -该位运算去除 n 的位级表示中最低的那一位。 +n&(n-1) 位运算可以将 n 的位级表示中最低的那一位 1 设置为 0。不断将 1 设置为 0，直到 n 为 0。时间复杂度：O(M)，其中 M 表示 1 的个数。 -``` -n : 10110100 -n-1 : 10110011 -n&(n-1) : 10110000 -``` - -时间复杂度：O(M)，其中 M 表示 1 的个数。 +

```java @@ -31,15 +27,6 @@ public int NumberOf1(int n) { ``` -### Integer.bitCount() - -```java -public int NumberOf1(int n) { - return Integer.bitCount(n); -} -``` - - diff --git a/notes/15. 二进制中 1 的个数.md b/notes/15. 二进制中 1 的个数.md index 5afa17a2..d3532e14 100644 --- a/notes/15. 二进制中 1 的个数.md +++ b/notes/15. 二进制中 1 的个数.md @@ -1,22 +1,18 @@ # 15. 二进制中 1 的个数 -[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) +## 题目链接 + +[牛客网](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) ## 题目描述 输入一个整数，输出该数二进制表示中 1 的个数。 -### n&(n-1) +### 解题思路 -该位运算去除 n 的位级表示中最低的那一位。 +n&(n-1) 位运算可以将 n 的位级表示中最低的那一位 1 设置为 0。不断将 1 设置为 0，直到 n 为 0。时间复杂度：O(M)，其中 M 表示 1 的个数。 -``` -n : 10110100 -n-1 : 10110011 -n&(n-1) : 10110000 -``` - -时间复杂度：O(M)，其中 M 表示 1 的个数。 +

```java @@ -31,15 +27,6 @@ public int NumberOf1(int n) { ``` -### Integer.bitCount() - -```java -public int NumberOf1(int n) { - return Integer.bitCount(n); -} -``` - - diff --git a/notes/pics/image-20201105004127554.png b/notes/pics/image-20201105004127554.png new file mode 100644 index 0000000000000000000000000000000000000000..8c39b9e21955dd6df699519eca481e073c2373a5 GIT binary patch literal 14598 zcmeHuXH=72yCyd5@Cv9%wFHo^NRz4}(mM&gC>^Ae(1{HZ5NXnzbO=pKD4~g&3G-Bg5oNL!o&O89*Ik%-cE-2q*-FuyaQSv*Kbq;G5N^G zcW~LX-i!L0s&Bb2EPw%AjI}G-O`$a59=Wpe;LZxts ztuB+!+q15IBHT%;*(bo%C@7h}IFR3;?F3vs0^UE--#<$J^Y)A5FfvhCKQ0)}t`;o9WyC{vg>7{~F`O=KVNN@zqQZR-6C~-^}bjqOr){Zc|jV zu2`J#tnHpy5+8y#>lMpyH|1~o1WM?tyGYU`CQqhI5pqH$ab}^rKZeU}A|dnhtOsc= zym1N9CIdWy8&=m1O$ajvzH{OKnHOR9{CJ9d)l?Oo#v2cvNPC!v2n<^N$Y6h0B5 zh~4-cP_|LWI(N^o2W!&9kF3!M`Qo5e%+@Hj^VQXB8OhYwo&52aOIx*1i164R+vvu( z6DCQ;qR2&!l(WvhDn?WGMBC;rJnTWwMfZhjlPXy_`Qy$9KhmX2gIG>~QY|qk@^iGB 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