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feat: add solutions to lc problem: No.0880 (doocs#2014)
No.0880.Decoded String at Index
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lines changed

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‎solution/0800-0899/0880.Decoded String at Index/README.md

Lines changed: 127 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -59,22 +59,148 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：逆向思维**
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我们可以先计算出解码字符串的总长度 $m$，然后从后向前遍历字符串，每次更新 $k$ 为 $k \bmod m$，直到 $k$ 为 $0$ 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 $1$。
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时间复杂度 $O(n)$，其中 $n$ 为字符串的长度。空间复杂度 $O(1)$。
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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 decodeAtIndex(self, s: str, k: int) -> str:
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m = 0
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for c in s:
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if c.isdigit():
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m *= int(c)
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else:
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m += 1
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for c in s[::-1]:
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k %= m
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if k == 0 and c.isalpha():
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return c
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if c.isdigit():
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m //= int(c)
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else:
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m -= 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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public String decodeAtIndex(String s, int k) {
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long m = 0;
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for (int i = 0; i < s.length(); ++i) {
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if (Character.isDigit(s.charAt(i))) {
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m *= (s.charAt(i) - '0');
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} else {
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++m;
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}
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}
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for (int i = s.length() - 1;; --i) {
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k %= m;
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if (k == 0 && !Character.isDigit(s.charAt(i))) {
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return String.valueOf(s.charAt(i));
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}
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if (Character.isDigit(s.charAt(i))) {
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m /= (s.charAt(i) - '0');
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} else {
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--m;
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}
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}
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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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string decodeAtIndex(string s, int k) {
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long long m = 0;
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for (char& c : s) {
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if (isdigit(c)) {
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m *= (c - '0');
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} else {
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++m;
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}
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}
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for (int i = s.size() - 1;; --i) {
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k %= m;
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if (k == 0 && isalpha(s[i])) {
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return string(1, s[i]);
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}
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if (isdigit(s[i])) {
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m /= (s[i] - '0');
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} else {
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--m;
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}
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}
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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 decodeAtIndex(s string, k int) string {
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m := 0
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for _, c := range s {
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if c >= '0' && c <= '9' {
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m *= int(c - '0')
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} else {
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m++
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}
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}
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for i := len(s) - 1; ; i-- {
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k %= m
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if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
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return string(s[i])
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}
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if s[i] >= '0' && s[i] <= '9' {
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m /= int(s[i] - '0')
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} else {
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m--
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}
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}
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}
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```
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### **TypeScript**
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```ts
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function decodeAtIndex(s: string, k: number): string {
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let m = 0n;
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for (const c of s) {
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if (c >= '1' && c <= '9') {
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m *= BigInt(c);
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} else {
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++m;
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}
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}
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for (let i = s.length - 1; ; --i) {
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if (k >= m) {
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k %= Number(m);
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}
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if (k === 0 && s[i] >= 'a' && s[i] <= 'z') {
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return s[i];
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}
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if (s[i] >= '1' && s[i] <= '9') {
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m = (m / BigInt(s[i])) | 0n;
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} else {
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--m;
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}
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}
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}
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```
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### **...**

‎solution/0800-0899/0880.Decoded String at Index/README_EN.md

Lines changed: 127 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -55,18 +55,144 @@ The 1<sup>st</sup> letter is &quot;a&quot;.
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## Solutions
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**Solution 1: Reverse Thinking**
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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 $k \bmod m$, until $k$ is $0$ 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 $1$ from $m$.
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The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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```python
63-
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class Solution:
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def decodeAtIndex(self, s: str, k: int) -> str:
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m = 0
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for c in s:
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if c.isdigit():
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m *= int(c)
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else:
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m += 1
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for c in s[::-1]:
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k %= m
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if k == 0 and c.isalpha():
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return c
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if c.isdigit():
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m //= int(c)
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else:
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m -= 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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public String decodeAtIndex(String s, int k) {
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long m = 0;
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for (int i = 0; i < s.length(); ++i) {
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if (Character.isDigit(s.charAt(i))) {
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m *= (s.charAt(i) - '0');
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} else {
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++m;
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}
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}
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for (int i = s.length() - 1;; --i) {
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k %= m;
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if (k == 0 && !Character.isDigit(s.charAt(i))) {
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return String.valueOf(s.charAt(i));
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}
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if (Character.isDigit(s.charAt(i))) {
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m /= (s.charAt(i) - '0');
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} else {
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--m;
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}
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}
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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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string decodeAtIndex(string s, int k) {
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long long m = 0;
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for (char& c : s) {
123+
if (isdigit(c)) {
124+
m *= (c - '0');
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} else {
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++m;
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}
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}
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for (int i = s.size() - 1;; --i) {
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k %= m;
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if (k == 0 && isalpha(s[i])) {
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return string(1, s[i]);
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}
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if (isdigit(s[i])) {
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m /= (s[i] - '0');
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} else {
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--m;
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}
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}
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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 decodeAtIndex(s string, k int) string {
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m := 0
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for _, c := range s {
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if c >= '0' && c <= '9' {
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m *= int(c - '0')
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} else {
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m++
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}
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}
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for i := len(s) - 1; ; i-- {
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k %= m
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if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
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return string(s[i])
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}
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if s[i] >= '0' && s[i] <= '9' {
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m /= int(s[i] - '0')
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} else {
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m--
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}
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}
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}
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```
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### **TypeScript**
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```ts
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function decodeAtIndex(s: string, k: number): string {
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let m = 0n;
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for (const c of s) {
176+
if (c >= '1' && c <= '9') {
177+
m *= BigInt(c);
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} else {
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++m;
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}
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}
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for (let i = s.length - 1; ; --i) {
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if (k >= m) {
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k %= Number(m);
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}
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if (k === 0 && s[i] >= 'a' && s[i] <= 'z') {
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return s[i];
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}
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if (s[i] >= '1' && s[i] <= '9') {
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m = (m / BigInt(s[i])) | 0n;
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} else {
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--m;
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}
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}
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}
70196
```
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### **...**

‎solution/0800-0899/0880.Decoded String at Index/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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class Solution {
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public:
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string decodeAtIndex(string s, int k) {
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long long m = 0;
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for (char& c : s) {
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if (isdigit(c)) {
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m *= (c - '0');
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} else {
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++m;
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}
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}
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for (int i = s.size() - 1;; --i) {
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k %= m;
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if (k == 0 && isalpha(s[i])) {
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return string(1, s[i]);
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}
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if (isdigit(s[i])) {
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m /= (s[i] - '0');
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} else {
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--m;
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}
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}
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}
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};

‎solution/0800-0899/0880.Decoded String at Index/Solution.go

Lines changed: 21 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
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func decodeAtIndex(s string, k int) string {
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m := 0
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for _, c := range s {
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if c >= '0' && c <= '9' {
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m *= int(c - '0')
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} else {
7+
m++
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}
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}
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for i := len(s) - 1; ; i-- {
11+
k %= m
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if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
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return string(s[i])
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}
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if s[i] >= '0' && s[i] <= '9' {
16+
m /= int(s[i] - '0')
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} else {
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m--
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}
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}
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}

‎solution/0800-0899/0880.Decoded String at Index/Solution.java

Lines changed: 23 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
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class Solution {
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public String decodeAtIndex(String s, int k) {
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long m = 0;
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for (int i = 0; i < s.length(); ++i) {
5+
if (Character.isDigit(s.charAt(i))) {
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m *= (s.charAt(i) - '0');
7+
} else {
8+
++m;
9+
}
10+
}
11+
for (int i = s.length() - 1;; --i) {
12+
k %= m;
13+
if (k == 0 && !Character.isDigit(s.charAt(i))) {
14+
return String.valueOf(s.charAt(i));
15+
}
16+
if (Character.isDigit(s.charAt(i))) {
17+
m /= (s.charAt(i) - '0');
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} else {
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--m;
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}
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}
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}
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}

‎solution/0800-0899/0880.Decoded String at Index/Solution.py

Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution:
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def decodeAtIndex(self, s: str, k: int) -> str:
3+
m = 0
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for c in s:
5+
if c.isdigit():
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m *= int(c)
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else:
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m += 1
9+
for c in s[::-1]:
10+
k %= m
11+
if k == 0 and c.isalpha():
12+
return c
13+
if c.isdigit():
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m //= int(c)
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else:
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m -= 1

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