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Commit cf574fc

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feat: add python and java solutions to leetcode problem: No.1461
1461. Check If a String Contains All Binary Codes of Size K
1 parent f7eb6f8 commit cf574fc

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4 files changed

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-6
lines changed

4 files changed

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‎solution/1400-1499/1461.Check If a String Contains All Binary Codes of Size K/README.md‎

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<strong>输出:</strong>false
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<!-- 这里可写通用的实现逻辑 -->
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遍历字符串 s,用一个 set 存储所有长度为 k 的不同子串。只需要判断子串数能否达到 2<sup>k</sup> 即可。
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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 hasAllCodes(self, s: str, k: int) -> bool:
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counter = 1 << k
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exists = set()
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for i in range(k, len(s) + 1):
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if s[i - k: i] not in exists:
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exists.add(s[i - k: i])
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counter -= 1
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if counter == 0:
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return True
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return False
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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 boolean hasAllCodes(String s, int k) {
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int counter = 1 << k;
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Set<String> exists = new HashSet<>();
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for (int i = k; i <= s.length(); ++i) {
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String t = s.substring(i - k, i);
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if (!exists.contains(t)) {
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exists.add(t);
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--counter;
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}
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if (counter == 0) {
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return true;
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}
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}
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return false;
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}
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}
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```
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### **...**

‎solution/1400-1499/1461.Check If a String Contains All Binary Codes of Size K/README_EN.md‎

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@@ -63,13 +63,39 @@
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### **Python3**
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```python
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class Solution:
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def hasAllCodes(self, s: str, k: int) -> bool:
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counter = 1 << k
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exists = set()
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for i in range(k, len(s) + 1):
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if s[i - k: i] not in exists:
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exists.add(s[i - k: i])
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counter -= 1
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if counter == 0:
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return True
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return False
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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 boolean hasAllCodes(String s, int k) {
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int counter = 1 << k;
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Set<String> exists = new HashSet<>();
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for (int i = k; i <= s.length(); ++i) {
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String t = s.substring(i - k, i);
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if (!exists.contains(t)) {
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exists.add(t);
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--counter;
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}
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if (counter == 0) {
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return true;
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}
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}
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return false;
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}
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}
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```
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### **...**
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class Solution {
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public boolean hasAllCodes(String s, int k) {
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int counter = 1 << k;
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Set<String> exists = new HashSet<>();
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for (int i = k; i <= s.length(); ++i) {
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String t = s.substring(i - k, i);
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if (!exists.contains(t)) {
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exists.add(t);
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--counter;
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}
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if (counter == 0) {
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return true;
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}
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}
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return false;
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}
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}
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class Solution:
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def hasAllCodes(self, s: str, k: int) -> bool:
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counter = 1 << k
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exists = set()
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for i in range(k, len(s) + 1):
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if s[i - k: i] not in exists:
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exists.add(s[i - k: i])
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counter -= 1
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if counter == 0:
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return True
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return False

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