字节算法题_字节算法题库-优快云博客

本文链接：https://blog.youkuaiyun.com/zdfunk/article/details/119689626

本文介绍了将手机号转换为Base64编码的Java实现，并提供了使用滑动窗口查找特定长度子串的方法。通过对给定字符串进行遍历，检查是否能找到包含特定字符集合的连续子串，满足条件时返回子串起始位置。

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文章目录

- 手机号转base64编码
- 字符串找子串

手机号转base64编码

 public static void main(String[] args) {
        getBase64 getbase64 = new getBase64();
        long i = Long.parseLong("100");
        String str = getbase64.telToBase64(i);
        System.out.println( str);
    }

}

class getBase64{
    public String telToBase64(long lng){
        String base64[] =  {"A","B","C","D","E","F","G","H","I","J","K","L","M","N",
                "O","P","Q","R","S","T","U","V","W","X","Y","Z",
                "a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p",
                "q","r","s","t","u","v","w","x","y","z",
                "0","1","2","3","4","5","6","7","8","9","+","/"};
        String str = new String();
        String res= new String();
        int i,j;
        i = (int)(lng/64);
        j = (int)(lng%64);
        res=base64[j];
        while(i>0){
            j = i%64;
            i = i/64;
            str = base64[j];
            res= str + res;
        }
        return res;
    }
}

输出：Bk

100=64（B）*1+36（k）
在这里插入图片描述

字符串找子串

给定m个不重复的字符 [a, b, c, d]，以及一个长度为n的字符串tbcacbdata，问能否在这个字符串中找到一个长度为m的连续子串，使得这个子串刚好由上面m个字符组成，顺序无所谓，返回任意满足条件的一个子串的起始位置，未找到返回-1。比如上面这个例子，acbd，3。本题的子串需要满足长度为m，字符不重复，可以使用长为m的滑动窗口遍历字符串，窗口内每个字符都要出现一次，如果符合条件，就返回窗口起始位置。

class Solution {
    public int checkInclusion(char[] ch,String s) {
        int n = s.length();
        int m =ch.length;
        if(m>n) return -1;
        for(int i=0;i<n-m;i++){
            if(checkEqual(ch,s.substring(i,i+m))){
                return i;
            }
        }
        return -1;
    }
    public boolean checkEqual(char[] ch,String s){
        for(int i=0;i<ch.length;i++){
            if(s.indexOf(ch[i])==-1){
                return false;
            }
        }
        return true;
    }