/**
* 给定两个字符串 s1 和 s2,请编写一个程序,确定其中一个字符串的字符重新排列后,能否变成另一个字符串。
*
* 示例 1:
*
* 输入: s1 = "abc", s2 = "bca"
* 输出: true
* 示例 2:
*
* 输入: s1 = "abc", s2 = "bad"
* 输出: false
* 说明:
*
* 0 <= len(s1) <= 100
* 0 <= len(s2) <= 100
*
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/check-permutation-lcci
* 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
*/
public class 判定是否互为字符重排 {
public static boolean CheckPermutation(String s1, String s2) {
char[] s1Array = s1.toCharArray();
char[] s2Array = s2.toCharArray();
Arrays.sort(s1Array);
Arrays.sort(s2Array);
if (s1Array.length != s2Array.length) return false;
for (int i = 0; i < s1Array.length; i++) {
if (s1Array[i] != s2Array[i]) {
return false;
}
}
return true;
}
public static void tets( List<Integer> list){
if (list.isEmpty() )
System.out.println(list);
}
public static void main(String[] args) {
System.out.println(CheckPermutation("abc","bad"));
List<Integer> list = new ArrayList<>();
tets(list);
}
}