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feat: add solutions to lcci problem: No.05.04 (doocs#1388)
No.05.04.Closed Number
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‎lcci/05.04.Closed Number/README.md

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## 题目描述
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<!-- 这里写题目描述 -->
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<p>下一个数。给定一个正整数，找出与其二进制表达式中1的个数相同且大小最接近的那两个数（一个略大，一个略小）。</p>
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<p> <strong>示例1:</strong></p>
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<pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：位运算**
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我们先考虑如何找出第一个比 $num$ 大且二进制表示中 $1$ 的个数相同的数。
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我们可以从低位到高位遍历 $num$ 的相邻两个二进制位，如果低位为 $1$，且相邻的较高一位为 $0$，那么我们就找到了一个位置，我们可以将这个位置的 $0$ 变成 $1$，将这个位置的 $1$ 变成 $0$。然后我们把其余低位的 $1$ 全部移动到最低位，这样我们就得到了一个比 $num$ 大且二进制表示中 $1$ 的个数相同的数。
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同理，我们可以找到第一个比 $num$ 小且二进制表示中 $1$ 的个数相同的数。
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我们可以从低位到高位遍历 $num$ 的相邻两个二进制位，如果低位为 $0$，且相邻的较高一位为 $1$，那么我们就找到了一个位置，我们可以将这个位置的 $1$ 变成 $0$，将这个位置的 $0$ 变成 $1$。然后我们把其余低位的 $0$ 全部移动到最低位，这样我们就得到了一个比 $num$ 小且二进制表示中 $1$ 的个数相同的数。
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在实现上，我们可以用一段代码来统一处理以上两种情况。
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时间复杂度 $O(\log n)$，其中 $n$ 是 $num$ 的大小。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
35-
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class Solution:
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def findClosedNumbers(self, num: int) -> List[int]:
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ans = [-1] * 2
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dirs = (0, 1, 0)
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for p in range(2):
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x = num
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for i in range(1, 31):
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a, b = dirs[p], dirs[p + 1]
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if (x >> i & 1) == a and (x >> (i - 1) & 1) == b:
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x ^= 1 << i
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x ^= 1 << (i - 1)
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j, k = 0, i - 2
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while j < k:
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while j < k and (x >> j & 1) == b:
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j += 1
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while j < k and (x >> k & 1) == a:
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k -= 1
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if j < k:
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x ^= 1 << j
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x ^= 1 << k
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ans[p] = x
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break
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] findClosedNumbers(int num) {
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int[] ans = {-1, -1};
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int[] dirs = {0, 1, 0};
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for (int p = 0; p < 2; ++p) {
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int x = num;
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for (int i = 1; i < 31; ++i) {
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int a = dirs[p], b = dirs[p + 1];
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if ((x >> i & 1) == a && (x >> (i - 1) & 1) == b) {
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x ^= 1 << i;
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x ^= 1 << (i - 1);
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int j = 0, k = i - 2;
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while (j < k) {
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while (j < k && (x >> j & 1) == b) {
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++j;
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}
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while (j < k && (x >> k & 1) == a) {
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--k;
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}
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if (j < k) {
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x ^= 1 << j;
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x ^= 1 << k;
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}
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}
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ans[p] = x;
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break;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> findClosedNumbers(int num) {
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vector<int> ans(2, -1);
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int dirs[3] = {0, 1, 0};
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for (int p = 0; p < 2; ++p) {
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int x = num;
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for (int i = 1; i < 31; ++i) {
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int a = dirs[p], b = dirs[p + 1];
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if ((x >> i & 1) == a && (x >> (i - 1) & 1) == b) {
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x ^= 1 << i;
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x ^= 1 << (i - 1);
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int j = 0, k = i - 2;
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while (j < k) {
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while (j < k && (x >> j & 1) == b) {
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++j;
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}
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while (j < k && (x >> k & 1) == a) {
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--k;
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}
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if (j < k) {
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x ^= 1 << j;
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x ^= 1 << k;
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}
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}
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ans[p] = x;
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break;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findClosedNumbers(num int) []int {
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ans := []int{-1, -1}
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dirs := [3]int{0, 1, 0}
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for p := 0; p < 2; p++ {
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x := num
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for i := 1; i < 31; i++ {
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a, b := dirs[p], dirs[p+1]
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if x>>i&1 == a && x>>(i-1)&1 == b {
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x ^= 1 << i
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x ^= 1 << (i - 1)
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j, k := 0, i-2
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for j < k {
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for j < k && x>>j&1 == b {
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j++
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}
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for j < k && x>>k&1 == a {
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k--
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}
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if j < k {
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x ^= 1 << j
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x ^= 1 << k
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}
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}
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ans[p] = x
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break
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}
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}
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function findClosedNumbers(num: number): number[] {
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const ans: number[] = [-1, -1];
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const dirs: number[] = [0, 1, 0];
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for (let p = 0; p < 2; ++p) {
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let x = num;
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for (let i = 1; i < 31; ++i) {
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const [a, b] = [dirs[p], dirs[p + 1]];
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if (((x >> i) & 1) === a && ((x >> (i - 1)) & 1) === b) {
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x ^= 1 << i;
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x ^= 1 << (i - 1);
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let [j, k] = [0, i - 2];
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while (j < k) {
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while (j < k && ((x >> j) & 1) === b) {
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++j;
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}
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while (j < k && ((x >> k) & 1) === a) {
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--k;
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}
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if (j < k) {
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x ^= 1 << j;
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x ^= 1 << k;
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}
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}
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ans[p] = x;
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break;
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}
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}
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}
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return ans;
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}
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```
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### **...**

‎lcci/05.04.Closed Number/README_EN.md

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### **Python3**
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```python
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class Solution:
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def findClosedNumbers(self, num: int) -> List[int]:
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ans = [-1] * 2
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dirs = (0, 1, 0)
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for p in range(2):
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x = num
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for i in range(1, 31):
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a, b = dirs[p], dirs[p + 1]
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if (x >> i & 1) == a and (x >> (i - 1) & 1) == b:
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x ^= 1 << i
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x ^= 1 << (i - 1)
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j, k = 0, i - 2
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while j < k:
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while j < k and (x >> j & 1) == b:
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j += 1
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while j < k and (x >> k & 1) == a:
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k -= 1
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if j < k:
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x ^= 1 << j
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x ^= 1 << k
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ans[p] = x
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break
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int[] findClosedNumbers(int num) {
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int[] ans = {-1, -1};
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int[] dirs = {0, 1, 0};
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for (int p = 0; p < 2; ++p) {
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int x = num;
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for (int i = 1; i < 31; ++i) {
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int a = dirs[p], b = dirs[p + 1];
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if ((x >> i & 1) == a && (x >> (i - 1) & 1) == b) {
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x ^= 1 << i;
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x ^= 1 << (i - 1);
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int j = 0, k = i - 2;
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while (j < k) {
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while (j < k && (x >> j & 1) == b) {
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++j;
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}
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while (j < k && (x >> k & 1) == a) {
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--k;
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}
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if (j < k) {
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x ^= 1 << j;
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x ^= 1 << k;
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}
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}
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ans[p] = x;
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break;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> findClosedNumbers(int num) {
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vector<int> ans(2, -1);
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int dirs[3] = {0, 1, 0};
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for (int p = 0; p < 2; ++p) {
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int x = num;
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for (int i = 1; i < 31; ++i) {
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int a = dirs[p], b = dirs[p + 1];
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if ((x >> i & 1) == a && (x >> (i - 1) & 1) == b) {
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x ^= 1 << i;
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x ^= 1 << (i - 1);
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int j = 0, k = i - 2;
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while (j < k) {
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while (j < k && (x >> j & 1) == b) {
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++j;
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}
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while (j < k && (x >> k & 1) == a) {
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--k;
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}
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if (j < k) {
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x ^= 1 << j;
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x ^= 1 << k;
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}
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}
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ans[p] = x;
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break;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findClosedNumbers(num int) []int {
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ans := []int{-1, -1}
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dirs := [3]int{0, 1, 0}
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for p := 0; p < 2; p++ {
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x := num
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for i := 1; i < 31; i++ {
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a, b := dirs[p], dirs[p+1]
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if x>>i&1 == a && x>>(i-1)&1 == b {
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x ^= 1 << i
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x ^= 1 << (i - 1)
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j, k := 0, i-2
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for j < k {
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for j < k && x>>j&1 == b {
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j++
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}
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for j < k && x>>k&1 == a {
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k--
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}
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if j < k {
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x ^= 1 << j
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x ^= 1 << k
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}
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}
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ans[p] = x
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break
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}
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}
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function findClosedNumbers(num: number): number[] {
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const ans: number[] = [-1, -1];
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const dirs: number[] = [0, 1, 0];
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for (let p = 0; p < 2; ++p) {
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let x = num;
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for (let i = 1; i < 31; ++i) {
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const [a, b] = [dirs[p], dirs[p + 1]];
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if (((x >> i) & 1) === a && ((x >> (i - 1)) & 1) === b) {
183+
x ^= 1 << i;
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x ^= 1 << (i - 1);
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let [j, k] = [0, i - 2];
186+
while (j < k) {
187+
while (j < k && ((x >> j) & 1) === b) {
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++j;
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}
190+
while (j < k && ((x >> k) & 1) === a) {
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--k;
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}
193+
if (j < k) {
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x ^= 1 << j;
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x ^= 1 << k;
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}
197+
}
198+
ans[p] = x;
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break;
200+
}
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}
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}
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return ans;
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}
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```
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### **...**

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