字元串反轉案例

如何将字元串反轉

方法1 遞歸方法

// An highlighted block
 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2 通過 charAt(int index)傳回char值進行字元串拼接

複制代碼
 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3 把字元串轉換成字元數組倒叙拼接然後傳回值

複制代碼
 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4 調用StringBuffer中的reverse方法

public static String reverse4(String s) {
  return new StringBuffer(s).reverse().toString();
 }
           

//方法5 把字元串轉換成字元數組首位對調位置

複制代碼
 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6

複制代碼
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7

import java.util.Stack;
public class StringReverse {    
 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;
 } 
}
           

傳送門：【1】: https://www.cnblogs.com/lanseyitai1224/p/6955609.html