Article From:https://www.cnblogs.com/SkyeAngel/p/9060718.html

Reverse String II

Their own solution:

That is to use a StringBuffer to receive a string, using a pointer to indicate which part of the current K is, when I + K < s.length (), it is proved that the I can be reversed from the I to the I + k – 1 part.And from I + K part to I + 2 * k part is directly splicing, conditional J &lt, s.length () can be restricted, but forThe last small part of the restIf it is necessary to reverse, then you can not use the conditions used in the reverse above: from J = I + k – 1, but from j = s.length() – 1Start to the end of I

```class Solution {
public String reverseStr(String s, int k) {
if(s == null || s.length() == 0 || k == 0) return s;
StringBuffer str = new StringBuffer();
int i = 0;
while(i + k < s.length()){
for(int j = i + k - 1; j >= i && j < s.length(); j--){
str.append(s.charAt(j));
}
i += k;
for(int j = i; j < i + k && j < s.length(); j++){
str.append(s.charAt(j));
}
i += k;
}

if(i + k >= s.length()){
for(int j = s.length() - 1; j >= i; j--){
str.append(s.charAt(j));
}
}

return str.toString();
}
}
```

``` public String reverseStr(String s, int k) {
if(s == null || s.isEmpty()){return s;}
StringBuilder sb = new StringBuilder();
boolean reverse = true;
for(int i = 0; i < s.length(); i = i + k, reverse = !reverse){
int end = (i + k) < s.length() ? (i + k - 1) : s.length()-1;
if(reverse)
for(int j = end; j >= i; j--){sb.append(s.charAt(j));} // Copy in reverse
else
for(int j = i; j <= end; j++){sb.append(s.charAt(j));} // Copy as it is
}
return sb.toString();
}
```

```public String reverseStr(String s, int k) {
char[] ch = s.toCharArray();
for (int i = 0; i < ch.length; i += 2 * k) {
rev(ch, i, i + k);
}
return String.valueOf(ch);
}

private void rev(char[] ch, int i, int j) {
j = Math.min(ch.length, j) - 1;
for (; i < j; i++, j--) {
char tmp = ch[i];
ch[i] = ch[j];
ch[j] = tmp;
}
}
```