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Commit 42a5bae

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feat: add solutions to lc problem: No.0625
No.0625.Minimum Factorization
1 parent e030d3e commit 42a5bae

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6 files changed

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6 files changed

+227
-2
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‎solution/0600-0699/0625.Minimum Factorization/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:贪心 + 因式分解**
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我们先判断 $num$ 是否小于 2ドル,ドル如果是,直接返回 $num$。然后从 9ドル$ 开始,尽可能多地将数字分解为 9ドル,ドル然后分解为 8ドル,ドル以此类推,直到分解为 2ドル$。如果最后剩下的数字不是 1ドル,ドル或者结果超过了 2ドル^{31} - 1,ドル则返回 0ドル$。否则,我们返回结果。
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> 注意,分解后的数字,应该依次填充到结果的个位、十位、百位、千位......上,因此我们需要维护一个变量 $mul,ドル表示当前的位数。
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时间复杂度 $O(\log n),ドル空间复杂度 $O(1)$。其中 $n$ 为 $num$ 的值。
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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 smallestFactorization(self, num: int) -> int:
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if num < 2:
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return num
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ans, mul = 0, 1
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for i in range(9, 1, -1):
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while num % i == 0:
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num //= i
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ans = mul * i + ans
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mul *= 10
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return ans if num < 2 and ans <= 2**31 - 1 else 0
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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 smallestFactorization(int num) {
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if (num < 2) {
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return num;
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}
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long ans = 0, mul = 1;
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for (int i = 9; i >= 2; --i) {
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if (num % i == 0) {
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while (num % i == 0) {
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num /= i;
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ans = mul * i + ans;
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mul *= 10;
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}
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}
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}
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return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
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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 smallestFactorization(int num) {
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if (num < 2) {
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return num;
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}
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long long ans = 0, mul = 1;
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for (int i = 9; i >= 2; --i) {
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if (num % i == 0) {
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while (num % i == 0) {
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num /= i;
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ans = mul * i + ans;
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mul *= 10;
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}
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}
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}
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return num < 2 && ans <= INT_MAX ? ans : 0;
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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 smallestFactorization(num int) int {
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if num < 2 {
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return num
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}
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ans, mul := 0, 1
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for i := 9; i >= 2; i-- {
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if num%i == 0 {
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for num%i == 0 {
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num /= i
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ans = mul*i + ans
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mul *= 10
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}
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}
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}
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if num < 2 && ans <= math.MaxInt32 {
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return ans
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}
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return 0
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}
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```
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### **...**

‎solution/0600-0699/0625.Minimum Factorization/README_EN.md‎

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### **Python3**
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```python
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class Solution:
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def smallestFactorization(self, num: int) -> int:
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if num < 2:
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return num
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ans, mul = 0, 1
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for i in range(9, 1, -1):
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while num % i == 0:
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num //= i
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ans = mul * i + ans
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mul *= 10
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return ans if num < 2 and ans <= 2**31 - 1 else 0
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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 smallestFactorization(int num) {
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if (num < 2) {
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return num;
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}
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long ans = 0, mul = 1;
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for (int i = 9; i >= 2; --i) {
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if (num % i == 0) {
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while (num % i == 0) {
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num /= i;
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ans = mul * i + ans;
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mul *= 10;
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}
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}
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}
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return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
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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 smallestFactorization(int num) {
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if (num < 2) {
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return num;
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}
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long long ans = 0, mul = 1;
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for (int i = 9; i >= 2; --i) {
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if (num % i == 0) {
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while (num % i == 0) {
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num /= i;
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ans = mul * i + ans;
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mul *= 10;
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}
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}
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}
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return num < 2 && ans <= INT_MAX ? ans : 0;
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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 smallestFactorization(num int) int {
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if num < 2 {
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return num
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}
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ans, mul := 0, 1
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for i := 9; i >= 2; i-- {
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if num%i == 0 {
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for num%i == 0 {
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num /= i
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ans = mul*i + ans
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mul *= 10
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}
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}
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}
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if num < 2 && ans <= math.MaxInt32 {
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return ans
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}
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return 0
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}
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```
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### **...**
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class Solution {
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public:
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int smallestFactorization(int num) {
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if (num < 2) {
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return num;
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}
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long long ans = 0, mul = 1;
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for (int i = 9; i >= 2; --i) {
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if (num % i == 0) {
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while (num % i == 0) {
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num /= i;
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ans = mul * i + ans;
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mul *= 10;
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}
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}
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}
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return num < 2 && ans <= INT_MAX ? ans : 0;
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}
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};
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func smallestFactorization(num int) int {
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if num < 2 {
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return num
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}
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ans, mul := 0, 1
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for i := 9; i >= 2; i-- {
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if num%i == 0 {
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for num%i == 0 {
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num /= i
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ans = mul*i + ans
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mul *= 10
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}
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}
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}
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if num < 2 && ans <= math.MaxInt32 {
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return ans
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}
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return 0
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}
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class Solution {
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public int smallestFactorization(int num) {
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if (num < 2) {
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return num;
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}
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long ans = 0, mul = 1;
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for (int i = 9; i >= 2; --i) {
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if (num % i == 0) {
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while (num % i == 0) {
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num /= i;
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ans = mul * i + ans;
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mul *= 10;
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}
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}
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}
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return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
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}
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}
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@@ -0,0 +1,11 @@
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class Solution:
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def smallestFactorization(self, num: int) -> int:
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if num < 2:
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return num
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ans, mul = 0, 1
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for i in range(9, 1, -1):
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while num % i == 0:
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num //= i
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ans = mul * i + ans
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mul *= 10
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return ans if num < 2 and ans <= 2**31 - 1 else 0

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