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[pull] main from doocs:main #205

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pull merged 1 commit into FairyWorld:main from doocs:main
Mar 28, 2023
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85 changes: 84 additions & 1 deletion solution/0600-0699/0625.Minimum Factorization/README.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -42,22 +42,105 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:贪心 + 因式分解**

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。

> 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。

时间复杂度 $O(\log n),ドル空间复杂度 $O(1)$。其中 $n$ 为 $num$ 的值。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def smallestFactorization(self, num: int) -> int:
if num < 2:
return num
ans, mul = 0, 1
for i in range(9, 1, -1):
while num % i == 0:
num //= i
ans = mul * i + ans
mul *= 10
return ans if num < 2 and ans <= 2**31 - 1 else 0
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
}
}
```

### **C++**

```cpp
class Solution {
public:
int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= INT_MAX ? ans : 0;
}
};
```

### **Go**

```go
func smallestFactorization(num int) int {
if num < 2 {
return num
}
ans, mul := 0, 1
for i := 9; i >= 2; i-- {
if num%i == 0 {
for num%i == 0 {
num /= i
ans = mul*i + ans
mul *= 10
}
}
}
if num < 2 && ans <= math.MaxInt32 {
return ans
}
return 0
}
```

### **...**
Expand Down
77 changes: 76 additions & 1 deletion solution/0600-0699/0625.Minimum Factorization/README_EN.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -28,13 +28,88 @@
### **Python3**

```python

class Solution:
def smallestFactorization(self, num: int) -> int:
if num < 2:
return num
ans, mul = 0, 1
for i in range(9, 1, -1):
while num % i == 0:
num //= i
ans = mul * i + ans
mul *= 10
return ans if num < 2 and ans <= 2**31 - 1 else 0
```

### **Java**

```java
class Solution {
public int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
}
}
```

### **C++**

```cpp
class Solution {
public:
int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= INT_MAX ? ans : 0;
}
};
```

### **Go**

```go
func smallestFactorization(num int) int {
if num < 2 {
return num
}
ans, mul := 0, 1
for i := 9; i >= 2; i-- {
if num%i == 0 {
for num%i == 0 {
num /= i
ans = mul*i + ans
mul *= 10
}
}
}
if num < 2 && ans <= math.MaxInt32 {
return ans
}
return 0
}
```

### **...**
Expand Down
19 changes: 19 additions & 0 deletions solution/0600-0699/0625.Minimum Factorization/Solution.cpp
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
class Solution {
public:
int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= INT_MAX ? ans : 0;
}
};
19 changes: 19 additions & 0 deletions solution/0600-0699/0625.Minimum Factorization/Solution.go
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,19 @@
func smallestFactorization(num int) int {
if num < 2 {
return num
}
ans, mul := 0, 1
for i := 9; i >= 2; i-- {
if num%i == 0 {
for num%i == 0 {
num /= i
ans = mul*i + ans
mul *= 10
}
}
}
if num < 2 && ans <= math.MaxInt32 {
return ans
}
return 0
}
18 changes: 18 additions & 0 deletions solution/0600-0699/0625.Minimum Factorization/Solution.java
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
class Solution {
public int smallestFactorization(int num) {
if (num < 2) {
return num;
}
long ans = 0, mul = 1;
for (int i = 9; i >= 2; --i) {
if (num % i == 0) {
while (num % i == 0) {
num /= i;
ans = mul * i + ans;
mul *= 10;
}
}
}
return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0;
}
}
11 changes: 11 additions & 0 deletions solution/0600-0699/0625.Minimum Factorization/Solution.py
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,11 @@
class Solution:
def smallestFactorization(self, num: int) -> int:
if num < 2:
return num
ans, mul = 0, 1
for i in range(9, 1, -1):
while num % i == 0:
num //= i
ans = mul * i + ans
mul *= 10
return ans if num < 2 and ans <= 2**31 - 1 else 0

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