LeeCode_判定是否互为字符重排

题目

class Solution {
    public boolean CheckPermutation(String s1, String s2) {
        if(s1.length()!=s2.length()){
            return false;
        } 

        int[] array = new int[128];      # 存放字符对应count
        for (int i = 0; i < s1.length(); i++) {
            array[s1.charAt(i)]++;      # 返回index并++
            array[s2.charAt(i)]--;      # 返回index并--
        }

        for (int num : array) {
            if (num != 0) return false;      # 判断元素是否为0
        }
        return true;
    }
}
原文地址:https://www.cnblogs.com/Neil-J/p/13558462.html