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feat: add solutions to lc problem: No.1680
No.1680.Concatenation of Consecutive Binary Numbers
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‎solution/1600-1699/1680.Concatenation of Consecutive Binary Numbers/README.md

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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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观察数字的连接规律，我们可以发现，当连接到第 $i$ 个数时，实际上是将前 $i-1$ 个数连接而成的结果 $ans$ 往左移动一定的位数，然后再加上 $i$ 这个数，移动的位数 $shift$ 是 $i$ 中二进制的位数。由于 $i$ 在不断加 $1$，移动的位数要么与上一次移动的位数保持不变，要么加一。当 $i$ 为 $2$ 的幂次方的时候，也即是说 $i$ 的二进制数中只有一位是 $1$ 时，移动的位数相比于上次加 $1$。
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时间复杂度 $O(n)$，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
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class Solution:
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def concatenatedBinary(self, n: int) -> int:
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mod = 10**9 + 7
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ans = shift = 0
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for i in range(1, n + 1):
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if (i & (i - 1)) == 0:
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shift += 1
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ans = ((ans << shift) + i) % mod
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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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private static final int MOD = (int) 1e9 + 7;
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public int concatenatedBinary(int n) {
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long ans = 0;
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int shift = 0;
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for (int i = 1; i <= n; ++i) {
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if ((i & (i - 1)) == 0) {
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++shift;
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}
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ans = ((ans << shift) + i) % MOD;
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}
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return (int) 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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const int mod = 1e9 + 7;
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int concatenatedBinary(int n) {
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long ans = 0;
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int shift = 0;
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for (int i = 1; i <= n; ++i) {
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if ((i & (i - 1)) == 0) {
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++shift;
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}
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ans = ((ans << shift) + i) % mod;
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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 concatenatedBinary(n int) int {
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var ans, shift int
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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if i&(i-1) == 0 {
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shift++
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}
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ans = ((ans << shift) + i) % mod
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}
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return ans
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}
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```
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### **...**

‎solution/1600-1699/1680.Concatenation of Consecutive Binary Numbers/README_EN.md

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### **Python3**
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```python
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class Solution:
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def concatenatedBinary(self, n: int) -> int:
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mod = 10**9 + 7
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ans = shift = 0
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for i in range(1, n + 1):
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if (i & (i - 1)) == 0:
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shift += 1
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ans = ((ans << shift) + i) % mod
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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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private static final int MOD = (int) 1e9 + 7;
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public int concatenatedBinary(int n) {
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long ans = 0;
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int shift = 0;
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for (int i = 1; i <= n; ++i) {
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if ((i & (i - 1)) == 0) {
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++shift;
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}
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ans = ((ans << shift) + i) % MOD;
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}
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return (int) 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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const int mod = 1e9 + 7;
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int concatenatedBinary(int n) {
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long ans = 0;
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int shift = 0;
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for (int i = 1; i <= n; ++i) {
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if ((i & (i - 1)) == 0) {
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++shift;
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}
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ans = ((ans << shift) + i) % mod;
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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 concatenatedBinary(n int) int {
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var ans, shift int
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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if i&(i-1) == 0 {
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shift++
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}
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ans = ((ans << shift) + i) % mod
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}
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return ans
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}
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```
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### **...**

‎solution/1600-1699/1680.Concatenation of Consecutive Binary Numbers/Solution.cpp

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class Solution {
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public:
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const int mod = 1e9 + 7;
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int concatenatedBinary(int n) {
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long ans = 0;
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int shift = 0;
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for (int i = 1; i <= n; ++i) {
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if ((i & (i - 1)) == 0) {
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++shift;
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}
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ans = ((ans << shift) + i) % mod;
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}
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return ans;
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}
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};

‎solution/1600-1699/1680.Concatenation of Consecutive Binary Numbers/Solution.go

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func concatenatedBinary(n int) int {
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var ans, shift int
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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if i&(i-1) == 0 {
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shift++
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}
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ans = ((ans << shift) + i) % mod
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}
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return ans
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}

‎solution/1600-1699/1680.Concatenation of Consecutive Binary Numbers/Solution.java

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class Solution {
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private static final int MOD = (int) 1e9 + 7;
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public int concatenatedBinary(int n) {
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long ans = 0;
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int shift = 0;
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for (int i = 1; i <= n; ++i) {
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if ((i & (i - 1)) == 0) {
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++shift;
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}
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ans = ((ans << shift) + i) % MOD;
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}
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return (int) ans;
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}
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}

‎solution/1600-1699/1680.Concatenation of Consecutive Binary Numbers/Solution.py

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class Solution:
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def concatenatedBinary(self, n: int) -> int:
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mod = 10**9 + 7
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ans = shift = 0
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for i in range(1, n + 1):
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if (i & (i - 1)) == 0:
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shift += 1
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ans = ((ans << shift) + i) % mod
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return ans

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