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feat: add solutions to lc problem: No.0788
No.0788.Rotated Digits
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‎solution/0700-0799/0788.Rotated Digits/README.md

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@@ -35,22 +35,145 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：直接枚举**
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一种直观且有效的思路是，直接枚举 $[1, n]$ 中的每个数，判断其是否为好数，若为好数，则答案加一。
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那么题目的重点转化为如何判断一个数字 $x$ 是否为好数。判断的逻辑如下：
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我们先用哈希表 $d$ 记录每个有效数字对应的旋转数字，在这道题中，有效数字有 `0, 1, 8, 2, 5, 6, 9`，分别对应旋转数字 `0, 1, 8, 5, 2, 9, 6`。
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然后遍历数字 $x$ 的每一位数字 $v$，如果 $v$ 不在哈希表 $d$ 中，则说明 $x$ 不是好数，直接返回 `false`。否则，将数字 $v$ 对应的旋转数字 $d[v]$ 加入到 $y$ 中。最后，判断 $x$ 和 $y$ 是否相等，若不相等，则说明 $x$ 是好数，返回 `true`。
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时间复杂度 $O(n\times \log n)$。
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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 rotatedDigits(self, n: int) -> int:
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def check(x):
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y, t = 0, x
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k = 1
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while t:
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v = t % 10
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if v not in d:
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return False
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y = d[v] * k + y
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k *= 10
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t //= 10
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return x != y
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d = {0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
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return sum(check(i) for i in range(1, n + 1))
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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 Map<Integer, Integer> d = new HashMap<>();
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static {
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d.put(0, 0);
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d.put(1, 1);
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d.put(8, 8);
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d.put(2, 5);
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d.put(5, 2);
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d.put(6, 9);
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d.put(9, 6);
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}
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public int rotatedDigits(int n) {
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (check(i)) {
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++ans;
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}
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}
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return ans;
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}
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private boolean check(int x) {
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int y = 0, t = x;
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int k = 1;
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while (t > 0) {
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int v = t % 10;
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if (!d.containsKey(v)) {
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return false;
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}
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y = d.get(v) * k + y;
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k *= 10;
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t /= 10;
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}
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return x != y;
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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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unordered_map<int, int> d{{0, 0}, {1, 1}, {8, 8}, {2, 5}, {5, 2}, {6, 9}, {9, 6}};
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int rotatedDigits(int n) {
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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ans += check(i);
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}
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return ans;
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}
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bool check(int x) {
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int y = 0, t = x;
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int k = 1;
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while (t) {
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int v = t % 10;
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if (!d.count(v)) {
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return false;
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}
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y = d[v] * k + y;
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k *= 10;
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t /= 10;
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}
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return x != y;
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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 rotatedDigits(n int) int {
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d := map[int]int{0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
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check := func(x int) bool {
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y, t := 0, x
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k := 1
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for ; t > 0; t /= 10 {
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v := t % 10
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if _, ok := d[v]; !ok {
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return false
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}
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y = d[v]*k + y
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k *= 10
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}
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return x != y
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}
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ans := 0
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for i := 1; i <= n; i++ {
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if check(i) {
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ans++
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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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### **...**

‎solution/0700-0799/0788.Rotated Digits/README_EN.md

Lines changed: 112 additions & 1 deletion
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@@ -55,13 +55,124 @@ Note that 1 and 10 are not good numbers, since they remain unchanged after rotat
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### **Python3**
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```python
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class Solution:
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def rotatedDigits(self, n: int) -> int:
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def check(x):
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y, t = 0, x
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k = 1
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while t:
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v = t % 10
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if v not in d:
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return False
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y = d[v] * k + y
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k *= 10
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t //= 10
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return x != y
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d = {0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
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return sum(check(i) for i in range(1, n + 1))
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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 Map<Integer, Integer> d = new HashMap<>();
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static {
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d.put(0, 0);
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d.put(1, 1);
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d.put(8, 8);
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d.put(2, 5);
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d.put(5, 2);
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d.put(6, 9);
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d.put(9, 6);
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}
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public int rotatedDigits(int n) {
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (check(i)) {
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++ans;
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}
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}
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return ans;
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}
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private boolean check(int x) {
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int y = 0, t = x;
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int k = 1;
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while (t > 0) {
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int v = t % 10;
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if (!d.containsKey(v)) {
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return false;
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}
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y = d.get(v) * k + y;
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k *= 10;
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t /= 10;
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}
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return x != y;
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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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unordered_map<int, int> d{{0, 0}, {1, 1}, {8, 8}, {2, 5}, {5, 2}, {6, 9}, {9, 6}};
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int rotatedDigits(int n) {
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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ans += check(i);
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}
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return ans;
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}
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bool check(int x) {
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int y = 0, t = x;
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int k = 1;
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while (t) {
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int v = t % 10;
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if (!d.count(v)) {
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return false;
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}
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y = d[v] * k + y;
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k *= 10;
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t /= 10;
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}
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return x != y;
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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 rotatedDigits(n int) int {
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d := map[int]int{0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
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check := func(x int) bool {
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y, t := 0, x
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k := 1
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for ; t > 0; t /= 10 {
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v := t % 10
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if _, ok := d[v]; !ok {
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return false
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}
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y = d[v]*k + y
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k *= 10
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}
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return x != y
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}
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ans := 0
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for i := 1; i <= n; i++ {
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if check(i) {
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ans++
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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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### **...**

‎solution/0700-0799/0788.Rotated Digits/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,27 @@
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class Solution {
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public:
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unordered_map<int, int> d{{0, 0}, {1, 1}, {8, 8}, {2, 5}, {5, 2}, {6, 9}, {9, 6}};
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int rotatedDigits(int n) {
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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ans += check(i);
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}
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return ans;
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}
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bool check(int x) {
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int y = 0, t = x;
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int k = 1;
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while (t) {
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int v = t % 10;
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if (!d.count(v)) {
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return false;
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}
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y = d[v] * k + y;
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k *= 10;
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t /= 10;
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}
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return x != y;
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}
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};

‎solution/0700-0799/0788.Rotated Digits/Solution.go

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func rotatedDigits(n int) int {
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d := map[int]int{0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
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check := func(x int) bool {
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y, t := 0, x
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k := 1
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for ; t > 0; t /= 10 {
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v := t % 10
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if _, ok := d[v]; !ok {
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return false
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}
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y = d[v]*k + y
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k *= 10
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}
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return x != y
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}
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ans := 0
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for i := 1; i <= n; i++ {
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if check(i) {
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ans++
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}
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}
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return ans
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}

‎solution/0700-0799/0788.Rotated Digits/Solution.java

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@@ -0,0 +1,37 @@
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class Solution {
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private static final Map<Integer, Integer> d = new HashMap<>();
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static {
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d.put(0, 0);
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d.put(1, 1);
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d.put(8, 8);
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d.put(2, 5);
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d.put(5, 2);
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d.put(6, 9);
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d.put(9, 6);
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}
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public int rotatedDigits(int n) {
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (check(i)) {
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++ans;
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}
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}
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return ans;
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}
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private boolean check(int x) {
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int y = 0, t = x;
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int k = 1;
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while (t > 0) {
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int v = t % 10;
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if (!d.containsKey(v)) {
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return false;
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}
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y = d.get(v) * k + y;
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k *= 10;
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t /= 10;
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}
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return x != y;
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}
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}

‎solution/0700-0799/0788.Rotated Digits/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution:
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def rotatedDigits(self, n: int) -> int:
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def check(x):
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y, t = 0, x
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k = 1
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while t:
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v = t % 10
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if v not in d:
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return False
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y = d[v] * k + y
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k *= 10
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t //= 10
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return x != y
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d = {0: 0, 1: 1, 8: 8, 2: 5, 5: 2, 6: 9, 9: 6}
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return sum(check(i) for i in range(1, n + 1))

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