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Commit 14db2a5

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feat: add solutions to lc problem: No.1545 (#2071)
No.1545.Find Kth Bit in Nth Binary String
1 parent 6e71522 commit 14db2a5

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8 files changed

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‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README.md‎

Lines changed: 111 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -71,22 +71,132 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:分类讨论 + 递归**
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我们可以发现,对于 $S_n,ドル其前半部分和 $S_{n-1}$ 是一样的,而后半部分是 $S_{n-1}$ 的反转取反。因此我们可以设计一个函数 $dfs(n, k),ドル表示第 $n$ 个字符串的第 $k$ 位字符。答案即为 $dfs(n, k)$。
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函数 $dfs(n, k)$ 的计算过程如下:
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- 如果 $k = 1,ドル那么答案为 0ドル$;
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- 如果 $k$ 是 2ドル$ 的幂次方,那么答案为 1ドル$;
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- 如果 $k \times 2 \lt 2^n - 1,ドル说明 $k$ 在前半部分,答案为 $dfs(n - 1, k)$;
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- 否则,答案为 $dfs(n - 1, 2^n - k) \oplus 1,ドル其中 $\oplus$ 表示异或运算。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为题目给定的 $n$。
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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 findKthBit(self, n: int, k: int) -> str:
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def dfs(n: int, k: int) -> int:
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if k == 1:
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return 0
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if (k & (k - 1)) == 0:
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return 1
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m = 1 << n
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if k * 2 < m - 1:
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return dfs(n - 1, k)
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return dfs(n - 1, m - k) ^ 1
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return str(dfs(n, k))
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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 char findKthBit(int n, int k) {
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return (char) ('0' + dfs(n, k));
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}
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private int dfs(int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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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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char findKthBit(int n, int k) {
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function<int(int, int)> dfs = [&](int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return '0' + dfs(n, k);
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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 findKthBit(n int, k int) byte {
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var dfs func(n, k int) int
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dfs = func(n, k int) int {
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if k == 1 {
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return 0
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}
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if k&(k-1) == 0 {
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return 1
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}
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m := 1 << n
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if k*2 < m-1 {
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return dfs(n-1, k)
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}
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return dfs(n-1, m-k) ^ 1
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}
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return byte('0' + dfs(n, k))
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}
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```
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### **TypeScript**
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```ts
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function findKthBit(n: number, k: number): string {
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const dfs = (n: number, k: number): number => {
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if (k === 1) {
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return 0;
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}
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if ((k & (k - 1)) === 0) {
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return 1;
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}
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const m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return dfs(n, k).toString();
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}
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```
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### **...**

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -53,18 +53,128 @@ The 11<sup>th</sup> bit is &quot;1&quot;.
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## Solutions
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**Solution 1: Case Analysis + Recursion**
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We can observe that for $S_n,ドル the first half is the same as $S_{n-1},ドル and the second half is the reverse and negation of $S_{n-1}$. Therefore, we can design a function $dfs(n, k),ドル which represents the $k$-th character of the $n$-th string. The answer is $dfs(n, k)$.
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The calculation process of the function $dfs(n, k)$ is as follows:
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- If $k = 1,ドル then the answer is 0ドル$;
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- If $k$ is a power of 2ドル,ドル then the answer is 1ドル$;
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- If $k \times 2 < 2^n - 1,ドル it means that $k$ is in the first half, and the answer is $dfs(n - 1, k)$;
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- Otherwise, the answer is $dfs(n - 1, 2^n - k) \oplus 1,ドル where $\oplus$ represents the XOR operation.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the given $n$ in the problem.
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<!-- tabs:start -->
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### **Python3**
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```python
61-
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class Solution:
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def findKthBit(self, n: int, k: int) -> str:
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def dfs(n: int, k: int) -> int:
77+
if k == 1:
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return 0
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if (k & (k - 1)) == 0:
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return 1
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m = 1 << n
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if k * 2 < m - 1:
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return dfs(n - 1, k)
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return dfs(n - 1, m - k) ^ 1
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return str(dfs(n, k))
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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 char findKthBit(int n, int k) {
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return (char) ('0' + dfs(n, k));
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}
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private int dfs(int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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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 {
117+
public:
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char findKthBit(int n, int k) {
119+
function<int(int, int)> dfs = [&](int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return '0' + dfs(n, k);
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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 findKthBit(n int, k int) byte {
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var dfs func(n, k int) int
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dfs = func(n, k int) int {
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if k == 1 {
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return 0
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}
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if k&(k-1) == 0 {
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return 1
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}
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m := 1 << n
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if k*2 < m-1 {
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return dfs(n-1, k)
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}
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return dfs(n-1, m-k) ^ 1
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}
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return byte('0' + dfs(n, k))
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}
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```
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### **TypeScript**
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```ts
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function findKthBit(n: number, k: number): string {
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const dfs = (n: number, k: number): number => {
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if (k === 1) {
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return 0;
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}
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if ((k & (k - 1)) === 0) {
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return 1;
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}
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const m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return dfs(n, k).toString();
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}
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```
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### **...**
Lines changed: 19 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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class Solution {
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public:
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char findKthBit(int n, int k) {
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function<int(int, int)> dfs = [&](int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return '0' + dfs(n, k);
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}
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};
Lines changed: 17 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
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func findKthBit(n int, k int) byte {
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var dfs func(n, k int) int
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dfs = func(n, k int) int {
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if k == 1 {
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return 0
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}
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if k&(k-1) == 0 {
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return 1
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}
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m := 1 << n
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if k*2 < m-1 {
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return dfs(n-1, k)
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}
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return dfs(n-1, m-k) ^ 1
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}
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return byte('0' + dfs(n, k))
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}
Lines changed: 18 additions & 37 deletions
Original file line numberDiff line numberDiff line change
@@ -1,38 +1,19 @@
1-
class Solution {
2-
public char findKthBit(int n, int k) {
3-
if (k == 1 || n == 1) {
4-
return '0';
5-
}
6-
Set<Integer> set = new HashSet<>();
7-
int len = calcLength(n, set);
8-
if (set.contains(k)) {
9-
return '1';
10-
}
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// 中间,返回1
12-
if (k < len / 2) {
13-
return findKthBit(n - 1, k);
14-
} else {
15-
if (set.contains(len - k)) {
16-
return '1';
17-
}
18-
return r(findKthBit(n - 1, len - k + 1));
19-
}
20-
}
21-
22-
private char r(char b) {
23-
if (b == '0') {
24-
return '1';
25-
}
26-
return '0';
27-
}
28-
29-
private int calcLength(int n, Set<Integer> set) {
30-
if (n == 1) {
31-
return 1;
32-
}
33-
34-
int ans = 2 * calcLength(n - 1, set) + 1;
35-
set.add(ans + 1);
36-
return ans;
37-
}
1+
class Solution {
2+
public char findKthBit(int n, int k) {
3+
return (char) ('0' + dfs(n, k));
4+
}
5+
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private int dfs(int n, int k) {
7+
if (k == 1) {
8+
return 0;
9+
}
10+
if ((k & (k - 1)) == 0) {
11+
return 1;
12+
}
13+
int m = 1 << n;
14+
if (k * 2 < m - 1) {
15+
return dfs(n - 1, k);
16+
}
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return dfs(n - 1, m - k) ^ 1;
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}
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}
Lines changed: 13 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
class Solution:
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def findKthBit(self, n: int, k: int) -> str:
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def dfs(n: int, k: int) -> int:
4+
if k == 1:
5+
return 0
6+
if (k & (k - 1)) == 0:
7+
return 1
8+
m = 1 << n
9+
if k * 2 < m - 1:
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return dfs(n - 1, k)
11+
return dfs(n - 1, m - k) ^ 1
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13+
return str(dfs(n, k))
Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
1+
function findKthBit(n: number, k: number): string {
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const dfs = (n: number, k: number): number => {
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if (k === 1) {
4+
return 0;
5+
}
6+
if ((k & (k - 1)) === 0) {
7+
return 1;
8+
}
9+
const m = 1 << n;
10+
if (k * 2 < m - 1) {
11+
return dfs(n - 1, k);
12+
}
13+
return dfs(n - 1, m - k) ^ 1;
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};
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return dfs(n, k).toString();
16+
}

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