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yanglbmeyanglbme
committedOct 16, 2022
feat: add solutions to lc problems: No.2437~2440
* No.2437.Number of Valid Clock Times * No.2438.Range Product Queries of Powers * No.2439.Minimize Maximum of Array * No.2440.Create Components With Same Value
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‎solution/0300-0399/0336.Palindrome Pairs/README.md

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<strong>输出：</strong>[[0,1],[1,0]]
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</pre>
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<p><strong>提示：</strong></p>
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<ul>

‎solution/0500-0599/0554.Brick Wall/README.md

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<strong>输出：</strong>3
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</pre>
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<p><strong>提示：</strong></p>
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<ul>

‎solution/0700-0799/0775.Global and Local Inversions/README.md

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<strong>解释：</strong>有 2 个全局倒置，和 1 个局部倒置。
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</pre>
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<p><strong>提示：</strong></p>
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<ul>

‎solution/0800-0899/0830.Positions of Large Groups/README.md

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<strong>输出：</strong>[]
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</pre>
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<p><strong>提示：</strong></p>
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<ul>

‎solution/1400-1499/1409.Queries on a Permutation With Key/README_EN.md

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<li>In the beginning, you have the permutation <code>P=[1,2,3,...,m]</code>.</li>
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<li>For the current <code>i</code>, find the position of <code>queries[i]</code> in the permutation <code>P</code> (<strong>indexing from 0</strong>) and then move this at the beginning of the permutation <code>P.</code>&nbsp;Notice that the position of <code>queries[i]</code> in <code>P</code> is the result for <code>queries[i]</code>.</li>
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</ul>
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<p>Return an array containing the result for the given <code>queries</code>.</p>

‎solution/2400-2499/2437.Number of Valid Clock Times/README.md

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@@ -50,22 +50,104 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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直接枚举所有时间，判断是否匹配 `time` 字符串，如果匹配则计数加一。
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时间复杂度 $O(24 \times 60)$。
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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 countTime(self, time: str) -> int:
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def check(s, t):
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for a, b in zip(s, t):
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if a != b and b != '?':
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return 0
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return 1
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return sum(
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check(f'{h:02d}:{m:02d}', time) for h in range(24) for m in range(60)
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)
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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 countTime(String time) {
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int ans = 0;
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for (int h = 0; h < 24; ++h) {
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for (int m = 0; m < 60; ++m) {
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String s = String.format("%02d:%02d", h, m);
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int ok = 1;
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for (int i = 0; i < 5; ++i) {
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if (s.charAt(i) != time.charAt(i) && time.charAt(i) != '?') {
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ok = 0;
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break;
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}
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}
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ans += ok;
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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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int countTime(string time) {
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int ans = 0;
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for (int h = 0; h < 24; ++h) {
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for (int m = 0; m < 60; ++m) {
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char s[20];
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sprintf(s, "%02d:%02d", h, m);
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int ok = 1;
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for (int i = 0; i < 5; ++i) {
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if (s[i] != time[i] && time[i] != '?') {
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ok = 0;
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break;
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}
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}
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ans += ok;
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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 countTime(time string) int {
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ans := 0
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for h := 0; h < 24; h++ {
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for m := 0; m < 60; m++ {
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s := fmt.Sprintf("%02d:%02d", h, m)
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ok := 1
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for i := 0; i < 5; i++ {
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if s[i] != time[i] && time[i] != '?' {
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ok = 0
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break
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}
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}
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ans += ok
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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**

‎solution/2400-2499/2437.Number of Valid Clock Times/README_EN.md

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### **Python3**
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```python
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class Solution:
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def countTime(self, time: str) -> int:
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def check(s, t):
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for a, b in zip(s, t):
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if a != b and b != '?':
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return 0
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return 1
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return sum(
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check(f'{h:02d}:{m:02d}', time) for h in range(24) for m in range(60)
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)
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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 countTime(String time) {
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int ans = 0;
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for (int h = 0; h < 24; ++h) {
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for (int m = 0; m < 60; ++m) {
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String s = String.format("%02d:%02d", h, m);
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int ok = 1;
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for (int i = 0; i < 5; ++i) {
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if (s.charAt(i) != time.charAt(i) && time.charAt(i) != '?') {
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ok = 0;
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break;
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}
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}
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ans += ok;
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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
95+
class Solution {
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public:
97+
int countTime(string time) {
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int ans = 0;
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for (int h = 0; h < 24; ++h) {
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for (int m = 0; m < 60; ++m) {
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char s[20];
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sprintf(s, "%02d:%02d", h, m);
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int ok = 1;
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for (int i = 0; i < 5; ++i) {
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if (s[i] != time[i] && time[i] != '?') {
106+
ok = 0;
107+
break;
108+
}
109+
}
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ans += ok;
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}
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}
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return ans;
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}
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};
116+
```
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### **Go**
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```go
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func countTime(time string) int {
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ans := 0
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for h := 0; h < 24; h++ {
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for m := 0; m < 60; m++ {
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s := fmt.Sprintf("%02d:%02d", h, m)
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ok := 1
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for i := 0; i < 5; i++ {
128+
if s[i] != time[i] && time[i] != '?' {
129+
ok = 0
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break
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}
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}
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ans += ok
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}
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}
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return ans
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}
62138
```
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### **TypeScript**

‎solution/2400-2499/2437.Number of Valid Clock Times/Solution.cpp

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class Solution {
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public:
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int countTime(string time) {
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int ans = 0;
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for (int h = 0; h < 24; ++h) {
6+
for (int m = 0; m < 60; ++m) {
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char s[20];
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sprintf(s, "%02d:%02d", h, m);
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int ok = 1;
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for (int i = 0; i < 5; ++i) {
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if (s[i] != time[i] && time[i] != '?') {
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ok = 0;
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break;
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}
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}
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ans += ok;
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}
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}
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return ans;
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}
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};

‎solution/2400-2499/2437.Number of Valid Clock Times/Solution.go

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1+
func countTime(time string) int {
2+
ans := 0
3+
for h := 0; h < 24; h++ {
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for m := 0; m < 60; m++ {
5+
s := fmt.Sprintf("%02d:%02d", h, m)
6+
ok := 1
7+
for i := 0; i < 5; i++ {
8+
if s[i] != time[i] && time[i] != '?' {
9+
ok = 0
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break
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}
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}
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ans += ok
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}
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}
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return ans
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}

‎solution/2400-2499/2437.Number of Valid Clock Times/Solution.java

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1+
class Solution {
2+
public int countTime(String time) {
3+
int ans = 0;
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for (int h = 0; h < 24; ++h) {
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for (int m = 0; m < 60; ++m) {
6+
String s = String.format("%02d:%02d", h, m);
7+
int ok = 1;
8+
for (int i = 0; i < 5; ++i) {
9+
if (s.charAt(i) != time.charAt(i) && time.charAt(i) != '?') {
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ok = 0;
11+
break;
12+
}
13+
}
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ans += ok;
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}
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}
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return ans;
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}
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}

‎solution/2400-2499/2437.Number of Valid Clock Times/Solution.py

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class Solution:
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def countTime(self, time: str) -> int:
3+
def check(s, t):
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for a, b in zip(s, t):
5+
if a != b and b != '?':
6+
return 0
7+
return 1
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9+
return sum(
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check(f'{h:02d}:{m:02d}', time) for h in range(24) for m in range(60)
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)

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