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feat: add solutions to lc problem: No.0340
No.0340.Longest Substring with At Most K Distinct Characters
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‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：滑动窗口 + 哈希表**
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我们可以使用滑动窗口的思想，维护一个滑动窗口，使得窗口内的字符串中不同字符的个数不超过 $k$ 个。窗口内不同字符个数的统计可以用哈希表 `cnt` 来维护。
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我们使用两个指针 $j$ 和 $i$ 分别表示滑动窗口的左右边界。我们先移动右边界 $i$，将字符 $s[i]$ 加入到窗口内，扩大滑动窗口，若此时窗口内不同字符的个数超过 $k$ 个，则移动左边界 $j$，缩小滑动窗口，直到窗口内不同字符的个数不超过 $k$ 个。此时我们可以更新答案的最大值，即 $ans = max(ans, i - j + 1)$。
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时间复杂度 $O(n)$，空间复杂度 $O(\min(n, k))$。其中 $n$ 为字符串的长度。
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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 lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
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cnt = Counter()
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n = len(s)
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ans = j = 0
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for i, c in enumerate(s):
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cnt[c] += 1
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while len(cnt) > k:
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cnt[s[j]] -= 1
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if cnt[s[j]] == 0:
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cnt.pop(s[j])
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j += 1
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ans = max(ans, i - j + 1)
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return ans
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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 lengthOfLongestSubstringKDistinct(String s, int k) {
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Map<Character, Integer> cnt = new HashMap<>();
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int n = s.length();
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int ans = 0, j = 0;
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for (int i = 0; i < n; ++i) {
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char c = s.charAt(i);
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cnt.put(c, cnt.getOrDefault(c, 0) + 1);
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while (cnt.size() > k) {
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char t = s.charAt(j);
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cnt.put(t, cnt.getOrDefault(t, 0) - 1);
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if (cnt.get(t) == 0) {
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cnt.remove(t);
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}
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++j;
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}
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ans = Math.max(ans, i - j + 1);
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}
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return ans;
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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 lengthOfLongestSubstringKDistinct(string s, int k) {
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unordered_map<char, int> cnt;
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int n = s.size();
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int ans = 0, j = 0;
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for (int i = 0; i < n; ++i) {
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cnt[s[i]]++;
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while (cnt.size() > k) {
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if (--cnt[s[j]] == 0) {
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cnt.erase(s[j]);
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}
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++j;
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}
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ans = max(ans, i - j + 1);
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}
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return ans;
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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 lengthOfLongestSubstringKDistinct(s string, k int) (ans int) {
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cnt := map[byte]int{}
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j := 0
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for i := range s {
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cnt[s[i]]++
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for len(cnt) > k {
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cnt[s[j]]--
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if cnt[s[j]] == 0 {
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delete(cnt, s[j])
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}
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j++
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}
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ans = max(ans, i-j+1)
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/README_EN.md

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### **Python3**
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```python
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class Solution:
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def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
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cnt = Counter()
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n = len(s)
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ans = j = 0
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for i, c in enumerate(s):
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cnt[c] += 1
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while len(cnt) > k:
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cnt[s[j]] -= 1
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if cnt[s[j]] == 0:
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cnt.pop(s[j])
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j += 1
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ans = max(ans, i - j + 1)
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return ans
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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 lengthOfLongestSubstringKDistinct(String s, int k) {
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Map<Character, Integer> cnt = new HashMap<>();
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int n = s.length();
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int ans = 0, j = 0;
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for (int i = 0; i < n; ++i) {
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char c = s.charAt(i);
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cnt.put(c, cnt.getOrDefault(c, 0) + 1);
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while (cnt.size() > k) {
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char t = s.charAt(j);
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cnt.put(t, cnt.getOrDefault(t, 0) - 1);
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if (cnt.get(t) == 0) {
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cnt.remove(t);
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}
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++j;
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}
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ans = Math.max(ans, i - j + 1);
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}
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return ans;
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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 lengthOfLongestSubstringKDistinct(string s, int k) {
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unordered_map<char, int> cnt;
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int n = s.size();
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int ans = 0, j = 0;
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for (int i = 0; i < n; ++i) {
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cnt[s[i]]++;
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while (cnt.size() > k) {
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if (--cnt[s[j]] == 0) {
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cnt.erase(s[j]);
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}
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++j;
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}
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ans = max(ans, i - j + 1);
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}
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return ans;
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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 lengthOfLongestSubstringKDistinct(s string, k int) (ans int) {
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cnt := map[byte]int{}
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j := 0
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for i := range s {
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cnt[s[i]]++
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for len(cnt) > k {
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cnt[s[j]]--
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if cnt[s[j]] == 0 {
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delete(cnt, s[j])
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}
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j++
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}
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ans = max(ans, i-j+1)
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/Solution.cpp

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class Solution {
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public:
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int lengthOfLongestSubstringKDistinct(string s, int k) {
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unordered_map<char, int> cnt;
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int n = s.size();
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int ans = 0, j = 0;
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for (int i = 0; i < n; ++i) {
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cnt[s[i]]++;
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while (cnt.size() > k) {
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if (--cnt[s[j]] == 0) {
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cnt.erase(s[j]);
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}
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++j;
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}
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ans = max(ans, i - j + 1);
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}
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return ans;
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}
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};

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/Solution.go

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func lengthOfLongestSubstringKDistinct(s string, k int) (ans int) {
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cnt := map[byte]int{}
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j := 0
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for i := range s {
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cnt[s[i]]++
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for len(cnt) > k {
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cnt[s[j]]--
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if cnt[s[j]] == 0 {
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delete(cnt, s[j])
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}
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j++
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}
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ans = max(ans, i-j+1)
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/Solution.java

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class Solution {
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public int lengthOfLongestSubstringKDistinct(String s, int k) {
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Map<Character, Integer> cnt = new HashMap<>();
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int n = s.length();
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int ans = 0, j = 0;
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for (int i = 0; i < n; ++i) {
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char c = s.charAt(i);
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cnt.put(c, cnt.getOrDefault(c, 0) + 1);
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while (cnt.size() > k) {
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char t = s.charAt(j);
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cnt.put(t, cnt.getOrDefault(t, 0) - 1);
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if (cnt.get(t) == 0) {
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cnt.remove(t);
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}
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++j;
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}
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ans = Math.max(ans, i - j + 1);
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}
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return ans;
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}
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}

‎solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/Solution.py

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class Solution:
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def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
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cnt = Counter()
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n = len(s)
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ans = j = 0
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for i, c in enumerate(s):
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cnt[c] += 1
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while len(cnt) > k:
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cnt[s[j]] -= 1
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if cnt[s[j]] == 0:
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cnt.pop(s[j])
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j += 1
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ans = max(ans, i - j + 1)
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return ans

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