判断是否为字符串重排(力扣)
程序员文章站
2024-02-21 14:10:10
...
import java.util.Arrays;
public class Test2 {
public static void main(String[] args) {
String s1 = "abc";
String s2 = "cba";
String s3 = "bcd";
System.out.println(CheckPermutation(s1,s2));
System.out.println(CheckPermutation2(s1,s3));
}
//字符串长度相等且互相包含
public static boolean CheckPermutation(String s1, String s2) {
if (s1.length() != s2.length()){
return false;
}
for (int i = 0; i < s1.length(); i++) {
if (!s2.contains(String.valueOf(s1.charAt(i)))){
return false;
}
}
for (int i = 0; i < s2.length(); i++) {
if (!s1.contains(String.valueOf(s2.charAt(i)))){
return false;
}
}
return true;
}
//转字符数组排序再比较是否相等
public static boolean CheckPermutation2(String s1, String s2) {
char[] c1 = s1.toCharArray();
Arrays.sort(c1);
char[] c2 = s2.toCharArray();
Arrays.sort(c2);
return c1.equals(c2);
}
}
下一篇: 异常问题-异常处理