Java中字符串反转的七种方法

package test1;

import java.util.Stack;

public class Reverse {
//方法一：递归+字符串截取；不断的进行二分截取，递归返回；
/*
* 【时间复杂度：】渐渐跟下去就是n,n/2,n/4,....n/2^k，其中k就是循环的次数
由于你n/2^k取整后>=1
即令n/2^k=1
可得k=log2n,（是以2为底，n的对数）
所以时间复杂度可以表示O()=O(logn)
*/
public static String reverse1(String s) {
int length = s.length();
if (length <= 1)
return s;
String left = s.substring(0, length / 2);
String right = s.substring(length / 2, length);
return reverse1(right) + reverse1(left);
}
//方法二：字符串拼接
/*
* 【时间复杂度：】O()=O(n)
*/
public static String reverse2(String s) {
int length = s.length();
String reverse = "";
for (int i = 0; i < length; i++){
reverse = s.charAt(i) + reverse;
}
return reverse;
}
//方法三：字符串拼接同2,
/*
* 【时间复杂度：】O()=O(n)
*/
public static String reverse3(String s) {
char[] array = s.toCharArray();
String reverse = "";
for (int i = array.length - 1; i >= 0; i--)
reverse += array[i];

return reverse;
}
//方法四：StringBuffer(s).reverse()
/*
* 【时间复杂度：】O()=O(n)
*/
public static String reverse4(String s) {
return new StringBuffer(s).reverse().toString();
}
//方法五：二分交换
/*
* 【时间复杂度：】O()=O(n)
*/
public static String reverse5(String orig) {
char[] s = orig.toCharArray();
int n = s.length - 1;
int halfLength = n / 2;
for (int i = 0; i <= halfLength; i++) {
char temp = s[i];
s[i] = s[n - i];
s[n - i] = temp;
}
return new String(s);
}
//方法六：异或
/*
* 【时间复杂度：】O()=O(n)
*/
public static String reverse6(String s) {

char[] str = s.toCharArray();

int begin = 0;
int end = s.length() - 1;

while (begin < end) {
str[begin] = (char) (str[begin] ^ str[end]);
str[end] = (char) (str[begin] ^ str[end]);
str[begin] = (char) (str[end] ^ str[begin]);
begin++;
end--;
}

return new String(str);
}
//方法二：辅助空间栈
/*
* 【时间复杂度：】O()=O(n)
*/
public static String reverse7(String s) {
char[] str = s.toCharArray();
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < str.length; i++)
stack.push(str[i]);

String reversed = "";
for (int i = 0; i < str.length; i++)
reversed += stack.pop();

return reversed;
}
public static void main(String[] args) {
System.out.println(reverse2("abcdefg"));
}
}