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feat: add solutions to lc problem: No.1156
No.1156.Swap For Longest Repeated Character Substring
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‎solution/1100-1199/1156.Swap For Longest Repeated Character Substring/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：双指针**
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我们先统计每个字符出现的次数，记录在数组 `cnt` 中。
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然后我们使用双指针 $i$ 和 $j$，初始时 $i = j = 0$，然后我们不断地向右移动 $j$，直到 $j$ 指向的字符与 $i$ 指向的字符不同，此时我们得到了一个长度为 $l = j - i$ 的子串，其中所有字符都相同。然后我们跳过第 $j$ 个字符，用指针 $k$ 继续向右移动，直到 $k$ 指向的字符与 $i$ 指向的字符不同，此时我们得到了一个长度为 $r = k - j - 1$ 的子串，其中所有字符都相同。此时我们可以得到的最长子串长度为 $min(l + r + 1, cnt[text[i]])$，其中 $cnt[text[i]]$ 表示字符 $text[i]$ 出现的次数。我们将这个值与当前的最大值进行比较，取较大值作为答案。
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时间复杂度为 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 为字符串的长度，而 $C$ 为字符集的大小。本题中 $C = 26$。
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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 maxRepOpt1(self, text: str) -> int:
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cnt = Counter(text)
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n = len(text)
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ans = i = 0
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while i < n:
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j = i
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while j < n and text[j] == text[i]:
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j += 1
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l = j - i
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k = j + 1
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while k < n and text[k] == text[i]:
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k += 1
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r = k - j - 1
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ans = max(ans, min(l + r + 1, cnt[text[i]]))
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i = j
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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 maxRepOpt1(String text) {
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int[] cnt = new int[26];
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int n = text.length();
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for (int i = 0; i < n; ++i) {
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++cnt[text.charAt(i) - 'a'];
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}
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int ans = 0, i = 0;
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while (i < n) {
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int j = i;
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while (j < n && text.charAt(j) == text.charAt(i)) {
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++j;
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}
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int l = j - i;
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int k = j + 1;
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while (k < n && text.charAt(k) == text.charAt(i)) {
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++k;
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}
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int r = k - j - 1;
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ans = Math.max(ans, Math.min(l + r + 1, cnt[text.charAt(i) - 'a']));
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i = j;
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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 maxRepOpt1(string text) {
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int cnt[26] = {0};
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for (char& c : text) {
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++cnt[c - 'a'];
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}
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int n = text.size();
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int ans = 0, i = 0;
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while (i < n) {
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int j = i;
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while (j < n && text[j] == text[i]) {
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++j;
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}
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int l = j - i;
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int k = j + 1;
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while (k < n && text[k] == text[i]) {
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++k;
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}
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int r = k - j - 1;
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ans = max(ans, min(l + r + 1, cnt[text[i] - 'a']));
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i = j;
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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 maxRepOpt1(text string) (ans int) {
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cnt := [26]int{}
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for _, c := range text {
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cnt[c-'a']++
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}
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n := len(text)
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for i, j := 0, 0; i < n; i = j {
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j = i
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for j < n && text[j] == text[i] {
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j++
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}
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l := j - i
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k := j + 1
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for k < n && text[k] == text[i] {
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k++
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}
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r := k - j - 1
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ans = max(ans, min(l+r+1, cnt[text[i]-'a']))
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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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func min(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/1100-1199/1156.Swap For Longest Repeated Character Substring/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maxRepOpt1(self, text: str) -> int:
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cnt = Counter(text)
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n = len(text)
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ans = i = 0
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while i < n:
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j = i
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while j < n and text[j] == text[i]:
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j += 1
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l = j - i
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k = j + 1
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while k < n and text[k] == text[i]:
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k += 1
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r = k - j - 1
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ans = max(ans, min(l + r + 1, cnt[text[i]]))
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i = j
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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 maxRepOpt1(String text) {
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int[] cnt = new int[26];
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int n = text.length();
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for (int i = 0; i < n; ++i) {
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++cnt[text.charAt(i) - 'a'];
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}
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int ans = 0, i = 0;
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while (i < n) {
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int j = i;
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while (j < n && text.charAt(j) == text.charAt(i)) {
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++j;
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}
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int l = j - i;
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int k = j + 1;
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while (k < n && text.charAt(k) == text.charAt(i)) {
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++k;
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}
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int r = k - j - 1;
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ans = Math.max(ans, Math.min(l + r + 1, cnt[text.charAt(i) - 'a']));
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i = j;
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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 maxRepOpt1(string text) {
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int cnt[26] = {0};
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for (char& c : text) {
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++cnt[c - 'a'];
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}
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int n = text.size();
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int ans = 0, i = 0;
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while (i < n) {
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int j = i;
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while (j < n && text[j] == text[i]) {
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++j;
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}
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int l = j - i;
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int k = j + 1;
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while (k < n && text[k] == text[i]) {
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++k;
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}
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int r = k - j - 1;
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ans = max(ans, min(l + r + 1, cnt[text[i] - 'a']));
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i = j;
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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 maxRepOpt1(text string) (ans int) {
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cnt := [26]int{}
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for _, c := range text {
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cnt[c-'a']++
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}
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n := len(text)
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for i, j := 0, 0; i < n; i = j {
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j = i
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for j < n && text[j] == text[i] {
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j++
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}
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l := j - i
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k := j + 1
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for k < n && text[k] == text[i] {
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k++
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}
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r := k - j - 1
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ans = max(ans, min(l+r+1, cnt[text[i]-'a']))
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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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func min(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/1100-1199/1156.Swap For Longest Repeated Character Substring/Solution.cpp

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class Solution {
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public:
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int maxRepOpt1(string text) {
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int cnt[26] = {0};
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for (char& c : text) {
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++cnt[c - 'a'];
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}
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int n = text.size();
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int ans = 0, i = 0;
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while (i < n) {
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int j = i;
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while (j < n && text[j] == text[i]) {
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++j;
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}
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int l = j - i;
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int k = j + 1;
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while (k < n && text[k] == text[i]) {
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++k;
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}
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int r = k - j - 1;
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ans = max(ans, min(l + r + 1, cnt[text[i] - 'a']));
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i = j;
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}
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return ans;
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}
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};

‎solution/1100-1199/1156.Swap For Longest Repeated Character Substring/Solution.go

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func maxRepOpt1(text string) (ans int) {
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cnt := [26]int{}
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for _, c := range text {
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cnt[c-'a']++
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}
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n := len(text)
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for i, j := 0, 0; i < n; i = j {
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j = i
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for j < n && text[j] == text[i] {
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j++
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}
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l := j - i
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k := j + 1
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for k < n && text[k] == text[i] {
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k++
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}
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r := k - j - 1
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ans = max(ans, min(l+r+1, cnt[text[i]-'a']))
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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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func min(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/1100-1199/1156.Swap For Longest Repeated Character Substring/Solution.java

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class Solution {
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public int maxRepOpt1(String text) {
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int[] cnt = new int[26];
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int n = text.length();
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for (int i = 0; i < n; ++i) {
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++cnt[text.charAt(i) - 'a'];
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}
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int ans = 0, i = 0;
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while (i < n) {
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int j = i;
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while (j < n && text.charAt(j) == text.charAt(i)) {
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++j;
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}
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int l = j - i;
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int k = j + 1;
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while (k < n && text.charAt(k) == text.charAt(i)) {
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++k;
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}
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int r = k - j - 1;
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ans = Math.max(ans, Math.min(l + r + 1, cnt[text.charAt(i) - 'a']));
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i = j;
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}
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return ans;
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}
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}

‎solution/1100-1199/1156.Swap For Longest Repeated Character Substring/Solution.py

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class Solution:
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def maxRepOpt1(self, text: str) -> int:
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cnt = Counter(text)
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n = len(text)
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ans = i = 0
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while i < n:
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j = i
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while j < n and text[j] == text[i]:
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j += 1
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l = j - i
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k = j + 1
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while k < n and text[k] == text[i]:
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k += 1
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r = k - j - 1
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ans = max(ans, min(l + r + 1, cnt[text[i]]))
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i = j
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return ans

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