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feat: add solutions to lc problem: No.1100
No.1100.Find K-Length Substrings With No Repeated Characters
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‎solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/README.md

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Original file line numberDiff line numberDiff line change
@@ -39,22 +39,96 @@
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<!-- 这里可写通用的实现逻辑 -->
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固定大小的滑动窗口。
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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 numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
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ans = j = 0
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mp = {}
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for i, c in enumerate(s):
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if c in mp:
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j = max(j, mp[c] + 1)
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mp[c] = i
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if i - j + 1 >= k:
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ans += 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 numKLenSubstrNoRepeats(String s, int k) {
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int ans = 0;
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Map<Character, Integer> mp = new HashMap<>();
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for (int i = 0, j = 0; i < s.length(); ++i) {
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char c = s.charAt(i);
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if (mp.containsKey(c)) {
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j = Math.max(j, mp.get(c) + 1);
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}
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mp.put(c, i);
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if (i - j + 1 >= k) {
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++ans;
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}
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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 numKLenSubstrNoRepeats(string s, int k) {
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int ans = 0;
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unordered_map<int, int> mp;
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for (int i = 0, j = 0; i < s.size(); ++i)
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{
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char c = s[i];
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if (mp.count(c)) j = max(j, mp[c] + 1);
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mp[c] = i;
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if (i - j + 1 >= k) ++ans;
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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 numKLenSubstrNoRepeats(s string, k int) int {
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mp := make(map[rune]int)
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ans, j := 0, 0
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for i, c := range s {
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if v, ok := mp[c]; ok {
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j = max(j, v+1)
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}
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mp[c] = i
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if i-j+1 >= k {
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ans++
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}
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}
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return ans
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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/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/README_EN.md

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@@ -53,13 +53,85 @@ Notice K can be larger than the length of S. In this case is not possible to fin
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### **Python3**
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```python
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class Solution:
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def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
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ans = j = 0
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mp = {}
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for i, c in enumerate(s):
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if c in mp:
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j = max(j, mp[c] + 1)
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mp[c] = i
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if i - j + 1 >= k:
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ans += 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 numKLenSubstrNoRepeats(String s, int k) {
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int ans = 0;
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Map<Character, Integer> mp = new HashMap<>();
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for (int i = 0, j = 0; i < s.length(); ++i) {
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char c = s.charAt(i);
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if (mp.containsKey(c)) {
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j = Math.max(j, mp.get(c) + 1);
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}
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mp.put(c, i);
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if (i - j + 1 >= k) {
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++ans;
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}
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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 numKLenSubstrNoRepeats(string s, int k) {
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int ans = 0;
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unordered_map<int, int> mp;
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for (int i = 0, j = 0; i < s.size(); ++i)
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{
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char c = s[i];
102+
if (mp.count(c)) j = max(j, mp[c] + 1);
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mp[c] = i;
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if (i - j + 1 >= k) ++ans;
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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 numKLenSubstrNoRepeats(s string, k int) int {
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mp := make(map[rune]int)
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ans, j := 0, 0
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for i, c := range s {
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if v, ok := mp[c]; ok {
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j = max(j, v+1)
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}
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mp[c] = i
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if i-j+1 >= k {
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ans++
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}
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}
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return ans
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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/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.cpp

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class Solution {
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public:
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int numKLenSubstrNoRepeats(string s, int k) {
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int ans = 0;
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unordered_map<int, int> mp;
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for (int i = 0, j = 0; i < s.size(); ++i)
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{
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char c = s[i];
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if (mp.count(c)) j = max(j, mp[c] + 1);
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mp[c] = i;
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if (i - j + 1 >= k) ++ans;
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}
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return ans;
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}
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};

‎solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.go

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func numKLenSubstrNoRepeats(s string, k int) int {
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mp := make(map[rune]int)
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ans, j := 0, 0
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for i, c := range s {
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if v, ok := mp[c]; ok {
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j = max(j, v+1)
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}
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mp[c] = i
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if i-j+1 >= k {
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ans++
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}
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}
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return ans
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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/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.java

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class Solution {
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public int numKLenSubstrNoRepeats(String s, int k) {
3+
int ans = 0;
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Map<Character, Integer> mp = new HashMap<>();
5+
for (int i = 0, j = 0; i < s.length(); ++i) {
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char c = s.charAt(i);
7+
if (mp.containsKey(c)) {
8+
j = Math.max(j, mp.get(c) + 1);
9+
}
10+
mp.put(c, i);
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if (i - j + 1 >= k) {
12+
++ans;
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}
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}
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return ans;
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}
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}

‎solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.py

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class Solution:
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def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
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ans = j = 0
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mp = {}
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for i, c in enumerate(s):
6+
if c in mp:
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j = max(j, mp[c] + 1)
8+
mp[c] = i
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if i - j + 1 >= k:
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ans += 1
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return ans

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