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Jun 23, 2021
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✨refactor: Update 191、剑指 Offer 15 #21

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1 change: 0 additions & 1 deletion Index/位运算.md
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@@ -1,6 +1,5 @@
| 题目 | 题解 | 难度 | 推荐指数 |
| ------------------------------------------------------------ | ------------------------------------------------------------ | ---- | -------- |
| [90. 子集 II](https://leetcode-cn.com/problems/subsets-ii/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/subsets-ii/solution/gong-shui-san-xie-yi-ti-shuang-jie-hui-s-g77q/) | 中等 | 🤩🤩🤩🤩 |
| [137. 只出现一次的数字 II](https://leetcode-cn.com/problems/single-number-ii/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/single-number-ii/solution/gong-shui-san-xie-yi-ti-san-jie-ha-xi-bi-fku8/) | 中等 | 🤩🤩🤩 |
| [190. 颠倒二进制位](https://leetcode-cn.com/problems/reverse-bits/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/reverse-bits/solution/yi-ti-san-jie-dui-cheng-wei-zhu-wei-fen-ub1hi/) | 简单 | 🤩🤩🤩 |
| [191. 位1的个数](https://leetcode-cn.com/problems/number-of-1-bits/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/number-of-1-bits/solution/yi-ti-san-jie-wei-shu-jian-cha-you-yi-to-av1r/) | 简单 | 🤩🤩🤩 |
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12 changes: 6 additions & 6 deletions LeetCode/191-200/191. 位1的个数（简单）.md
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Expand Up @@ -41,10 +41,10 @@ Tag : 「位运算」

### 「位数检查」解法

![image.png](https://pic.leetcode-cn.com/1616375441-WGCssd-image.png)

一个朴素的做法是，对 `int` 的每一位进行检查，并统计 $1$ 的个数。

![image.png](https://pic.leetcode-cn.com/1616375441-WGCssd-image.png)

代码：
```Java []
public class Solution {
Expand All @@ -64,8 +64,6 @@ public class Solution {

### 「右移统计」解法

![image.png](https://pic.leetcode-cn.com/1616375636-fXCFNF-image.png)

对于方法一，即使 $n$ 的高位均为是 $0$，我们也会对此进行循环检查。

因此另外一个做法是：通过 `n & 1` 来统计当前 $n$ 的最低位是否为 $1$，同时每次直接对 $n$ 进行右移并高位补 0。
Expand All @@ -74,6 +72,8 @@ public class Solution {

这样的做法，可以确保只会循环到最高位的 $1$。

![image.png](https://pic.leetcode-cn.com/1616375636-fXCFNF-image.png)

代码：
```Java []
public class Solution {
Expand Down Expand Up @@ -124,10 +124,10 @@ public class Solution {

### 「分组统计」解法

![image.png](https://pic.leetcode-cn.com/1616378128-yBWadF-image.png)

以上三种解法都是 $O(k)$ 的，事实上我们可以通过分组统计的方式，做到比 $O(k)$ 更低的复杂度。

![image.png](https://pic.leetcode-cn.com/1616378128-yBWadF-image.png)

代码：
```Java []
public class Solution {
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14 changes: 7 additions & 7 deletions LeetCode/剑指 Offer/剑指 Offer 15. 二进制中1的个数（简单）.md
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Expand Up @@ -23,7 +23,7 @@ Tag : 「位运算」、「分治」
输入：11111111111111111111111111111101
输出：31
解释：输入的二进制串 11111111111111111111111111111101 中，共有 31 位为 '1'。
``` 
```

提示：
* 输入必须是长度为 32 的 二进制串 。
Expand All @@ -32,10 +32,10 @@ Tag : 「位运算」、「分治」

### 「位数检查」解法

![image.png](https://pic.leetcode-cn.com/1616375441-WGCssd-image.png)

一个朴素的做法是，对 `int` 的每一位进行检查，并统计 $1$ 的个数。

![image.png](https://pic.leetcode-cn.com/1616375441-WGCssd-image.png)

代码：
```Java []
public class Solution {
Expand All @@ -55,8 +55,6 @@ public class Solution {

### 「右移统计」解法

![image.png](https://pic.leetcode-cn.com/1616375636-fXCFNF-image.png)

对于方法一，即使 $n$ 的高位均为是 $0$，我们也会对此进行循环检查。

因此另外一个做法是：通过 `n & 1` 来统计当前 $n$ 的最低位是否为 $1$，同时每次直接对 $n$ 进行右移并高位补 0。
Expand All @@ -65,6 +63,8 @@ public class Solution {

这样的做法，可以确保只会循环到最高位的 $1$。

![image.png](https://pic.leetcode-cn.com/1616375636-fXCFNF-image.png)

代码：
```Java []
public class Solution {
Expand Down Expand Up @@ -115,10 +115,10 @@ public class Solution {

### 「分组统计」解法

![image.png](https://pic.leetcode-cn.com/1616378128-yBWadF-image.png)

以上三种解法都是 $O(k)$ 的，事实上我们可以通过分组统计的方式，做到比 $O(k)$ 更低的复杂度。

![image.png](https://pic.leetcode-cn.com/1616378128-yBWadF-image.png)

代码：
```Java []
public class Solution {
Expand Down