刷题笔记-面试

1.面试题01.01.判定字符是否唯一

https://leetcode-cn.com/problems/is-unique-lcci/

public boolean isUnique(String astr) {
        int mark=0;
        boolean flag = true;
        for(int i=0;i<astr.length();i++) {
            int lm = astr.charAt(i)-'a';
            if((mark & (1<<lm)) > 0) {
                flag = false;
                break;
            } else {
                mark |= (1<<lm);
            }
        }
        return flag;
    }

2.面试题 01.02. 判定是否互为字符重排

https://leetcode-cn.com/problems/check-permutation-lcci/

class Solution {
        public boolean CheckPermutation(String s1, String s2) {
        if(s1.length()!=s2.length())return false;
        Map<Character,Integer> mps1 = new HashMap<>();
        Map<Character,Integer> mps2 = new HashMap<>();
        for(int i=0;i<s1.length();i++) {
            if(mps1.get(s1.charAt(i)) == null) {
                mps1.put(s1.charAt(i),1);
            }else {
                mps1.put(s1.charAt(i),mps1.get(s1.charAt(i))+1);
            }
            if(mps2.get(s2.charAt(i)) == null) {
                mps2.put(s2.charAt(i),1);
            }else {
                mps2.put(s2.charAt(i),mps2.get(s2.charAt(i))+1);
            }
        }
        for(Map.Entry<Character,Integer> k:mps1.entrySet()){
            if(mps2.get(k.getKey()) != k.getValue())return false;
        }
        return true;
    }
}

面试题 01.03. URL化

一开始是直接尝试用字符串拼接，如果是空字符就拼接%20，否则就拼接原来的字符。但是超时。
题目描述

class Solution {
    public String replaceSpaces(String S, int length) {
        String res = "";
        int m=0;
        char[] s = S.toCharArray();
        
        for(int i=0;i<length;i++) {
            if(s[i] == ' '){
                m++;
            }
        }
        m = m*2+length;
        char[] arr = new char[m];
        int j=0;
        for(int i=0;i<length;i++) {
              if(s[i] == ' '){
                 arr[j++] = '%';
                 arr[j++] = '2';
                 arr[j++] = '0';   
              }else{
                 arr[j++] = s[i];
              }   
        }
        return new String(arr);
    }
}