判断一个字符串是否对称

/** * 判断一个字符串是否对称 */public class Symmetric {public static void main(String[] args) {String str = null;System.out.println(str + ":" + isSymmetric(str));str = "";System.out.println(str + ":" + isSymmetric(str));// 偶数长度str = "abccba";System.out.println(str + ":" + isSymmetric(str));// 奇数长度str = "abcdcba";System.out.println(str + ":" + isSymmetric(str));str = "abca";System.out.println(str + ":" + isSymmetric(str));}public static boolean isSymmetric(String str) {if (null == str) {return false;}for (int i = 0; i < str.length() / 2; i++) {// 比较距字符串两头长度相同的字符是否一样if (str.charAt(i) != str.charAt(str.length() - i - 1)) {return false;}}return true;}}

输出结果:

null:false:trueabccba:trueabcdcba:trueabca:false

public static void main(String[] args) {String str = "abcdcba";System.out.println(isHuiWen(str, 0, str.length() - 1));}public static boolean isHuiWen(String str, int index, int end){if(index > end || index == end){return true;}if(str.charAt(index) == str.charAt(end)){return isHuiWen(str, index + 1, end - 1);}return false;} 2 楼 Technoboy 2011-05-11 不是，叫回文，用一般用递归实现 3 楼 ljl_ss 2011-05-11 Technoboy 写道不是，叫回文，用一般用递归实现
个人看法递归只是代码看起来简洁,每递归一次只会增加JVM方法栈的长度,开销比博主的写法大