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feat: add solutions to lc problem: No.0880 #2014

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acbin merged 1 commit into main from dev-0880
Nov 25, 2023
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128 changes: 127 additions & 1 deletion solution/0800-0899/0880.Decoded String at Index/README.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -59,22 +59,148 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:逆向思维**

我们可以先计算出解码字符串的总长度 $m,ドル然后从后向前遍历字符串,每次更新 $k$ 为 $k \bmod m,ドル直到 $k$ 为 0ドル$ 且当前字符为字母,返回当前字符。否则,如果当前字符为数字,则将 $m$ 除以该数字。如果当前字符为字母,则将 $m$ 减 1ドル$。

时间复杂度 $O(n),ドル其中 $n$ 为字符串的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def decodeAtIndex(self, s: str, k: int) -> str:
m = 0
for c in s:
if c.isdigit():
m *= int(c)
else:
m += 1
for c in s[::-1]:
k %= m
if k == 0 and c.isalpha():
return c
if c.isdigit():
m //= int(c)
else:
m -= 1
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public String decodeAtIndex(String s, int k) {
long m = 0;
for (int i = 0; i < s.length(); ++i) {
if (Character.isDigit(s.charAt(i))) {
m *= (s.charAt(i) - '0');
} else {
++m;
}
}
for (int i = s.length() - 1;; --i) {
k %= m;
if (k == 0 && !Character.isDigit(s.charAt(i))) {
return String.valueOf(s.charAt(i));
}
if (Character.isDigit(s.charAt(i))) {
m /= (s.charAt(i) - '0');
} else {
--m;
}
}
}
}
```

### **C++**

```cpp
class Solution {
public:
string decodeAtIndex(string s, int k) {
long long m = 0;
for (char& c : s) {
if (isdigit(c)) {
m *= (c - '0');
} else {
++m;
}
}
for (int i = s.size() - 1;; --i) {
k %= m;
if (k == 0 && isalpha(s[i])) {
return string(1, s[i]);
}
if (isdigit(s[i])) {
m /= (s[i] - '0');
} else {
--m;
}
}
}
};
```

### **Go**

```go
func decodeAtIndex(s string, k int) string {
m := 0
for _, c := range s {
if c >= '0' && c <= '9' {
m *= int(c - '0')
} else {
m++
}
}
for i := len(s) - 1; ; i-- {
k %= m
if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
return string(s[i])
}
if s[i] >= '0' && s[i] <= '9' {
m /= int(s[i] - '0')
} else {
m--
}
}
}
```

### **TypeScript**

```ts
function decodeAtIndex(s: string, k: number): string {
let m = 0n;
for (const c of s) {
if (c >= '1' && c <= '9') {
m *= BigInt(c);
} else {
++m;
}
}
for (let i = s.length - 1; ; --i) {
if (k >= m) {
k %= Number(m);
}
if (k === 0 && s[i] >= 'a' && s[i] <= 'z') {
return s[i];
}
if (s[i] >= '1' && s[i] <= '9') {
m = (m / BigInt(s[i])) | 0n;
} else {
--m;
}
}
}
```

### **...**
Expand Down
128 changes: 127 additions & 1 deletion solution/0800-0899/0880.Decoded String at Index/README_EN.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -55,18 +55,144 @@ The 1<sup>st</sup> letter is &quot;a&quot;.

## Solutions

**Solution 1: Reverse Thinking**

We can first calculate the total length $m$ of the decoded string, then traverse the string from back to front. Each time, we update $k$ to be $k \bmod m,ドル until $k$ is 0ドル$ and the current character is a letter, then we return the current character. Otherwise, if the current character is a number, we divide $m$ by this number. If the current character is a letter, we subtract 1ドル$ from $m$.

The time complexity is $O(n),ドル where $n$ is the length of the string. The space complexity is $O(1)$.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def decodeAtIndex(self, s: str, k: int) -> str:
m = 0
for c in s:
if c.isdigit():
m *= int(c)
else:
m += 1
for c in s[::-1]:
k %= m
if k == 0 and c.isalpha():
return c
if c.isdigit():
m //= int(c)
else:
m -= 1
```

### **Java**

```java
class Solution {
public String decodeAtIndex(String s, int k) {
long m = 0;
for (int i = 0; i < s.length(); ++i) {
if (Character.isDigit(s.charAt(i))) {
m *= (s.charAt(i) - '0');
} else {
++m;
}
}
for (int i = s.length() - 1;; --i) {
k %= m;
if (k == 0 && !Character.isDigit(s.charAt(i))) {
return String.valueOf(s.charAt(i));
}
if (Character.isDigit(s.charAt(i))) {
m /= (s.charAt(i) - '0');
} else {
--m;
}
}
}
}
```

### **C++**

```cpp
class Solution {
public:
string decodeAtIndex(string s, int k) {
long long m = 0;
for (char& c : s) {
if (isdigit(c)) {
m *= (c - '0');
} else {
++m;
}
}
for (int i = s.size() - 1;; --i) {
k %= m;
if (k == 0 && isalpha(s[i])) {
return string(1, s[i]);
}
if (isdigit(s[i])) {
m /= (s[i] - '0');
} else {
--m;
}
}
}
};
```

### **Go**

```go
func decodeAtIndex(s string, k int) string {
m := 0
for _, c := range s {
if c >= '0' && c <= '9' {
m *= int(c - '0')
} else {
m++
}
}
for i := len(s) - 1; ; i-- {
k %= m
if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
return string(s[i])
}
if s[i] >= '0' && s[i] <= '9' {
m /= int(s[i] - '0')
} else {
m--
}
}
}
```

### **TypeScript**

```ts
function decodeAtIndex(s: string, k: number): string {
let m = 0n;
for (const c of s) {
if (c >= '1' && c <= '9') {
m *= BigInt(c);
} else {
++m;
}
}
for (let i = s.length - 1; ; --i) {
if (k >= m) {
k %= Number(m);
}
if (k === 0 && s[i] >= 'a' && s[i] <= 'z') {
return s[i];
}
if (s[i] >= '1' && s[i] <= '9') {
m = (m / BigInt(s[i])) | 0n;
} else {
--m;
}
}
}
```

### **...**
Expand Down
24 changes: 24 additions & 0 deletions solution/0800-0899/0880.Decoded String at Index/Solution.cpp
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
class Solution {
public:
string decodeAtIndex(string s, int k) {
long long m = 0;
for (char& c : s) {
if (isdigit(c)) {
m *= (c - '0');
} else {
++m;
}
}
for (int i = s.size() - 1;; --i) {
k %= m;
if (k == 0 && isalpha(s[i])) {
return string(1, s[i]);
}
if (isdigit(s[i])) {
m /= (s[i] - '0');
} else {
--m;
}
}
}
};
21 changes: 21 additions & 0 deletions solution/0800-0899/0880.Decoded String at Index/Solution.go
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
func decodeAtIndex(s string, k int) string {
m := 0
for _, c := range s {
if c >= '0' && c <= '9' {
m *= int(c - '0')
} else {
m++
}
}
for i := len(s) - 1; ; i-- {
k %= m
if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
return string(s[i])
}
if s[i] >= '0' && s[i] <= '9' {
m /= int(s[i] - '0')
} else {
m--
}
}
}
23 changes: 23 additions & 0 deletions solution/0800-0899/0880.Decoded String at Index/Solution.java
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
class Solution {
public String decodeAtIndex(String s, int k) {
long m = 0;
for (int i = 0; i < s.length(); ++i) {
if (Character.isDigit(s.charAt(i))) {
m *= (s.charAt(i) - '0');
} else {
++m;
}
}
for (int i = s.length() - 1;; --i) {
k %= m;
if (k == 0 && !Character.isDigit(s.charAt(i))) {
return String.valueOf(s.charAt(i));
}
if (Character.isDigit(s.charAt(i))) {
m /= (s.charAt(i) - '0');
} else {
--m;
}
}
}
}
16 changes: 16 additions & 0 deletions solution/0800-0899/0880.Decoded String at Index/Solution.py
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
class Solution:
def decodeAtIndex(self, s: str, k: int) -> str:
m = 0
for c in s:
if c.isdigit():
m *= int(c)
else:
m += 1
for c in s[::-1]:
k %= m
if k == 0 and c.isalpha():
return c
if c.isdigit():
m //= int(c)
else:
m -= 1
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