242. Valid Anagram
Posted Updated LeetCode / Easya minute read (About 163 words)

242. Valid Anagram

Problem

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Solution

两数组相等时，直接遍历两个数组并记录各个字符出现的数量。
一个数组遍历时用做加法，另一个做减法。
如果最后每个字符出现的数量均为0，则返回真。

Code

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class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length()!=t.length()){
            return false;
        }
        int[] dic = new int[26];
        for (int i = 0; i < s.length(); i++){
            dic[s.charAt(i)-'a']++;
            dic[t.charAt(i)-'a']--;
        }
        
        for(int num : dic){
            if ( num != 0 ){
                return false;
            }
        }
        return true;
    }
}
387. First Unique Character in a String
Posted Updated LeetCode / Easy2 minutes read (About 240 words)

387. First Unique Character in a String

Problems

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

Solution

遍历，数组统计记录出现次数。
如果数组未记录过，则将其index添加进列表中保存。

遍历列表，如果数组统计结果为1，则返回对应的index。
否则返回-1。

Code 1

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class Solution {
    public int firstUniqChar(String s) {
        List<Integer> arr = new ArrayList<>();
        int[] bin = new int[26];
        
        for(int i = 0; i < s.length(); i++){
            if(bin[s.charAt(i) - 'a'] == 0) arr.add(i);
            bin[s.charAt(i) - 'a']++;
        }
        
        for(int i = 0; i < arr.size(); i++){
            if(bin[s.charAt(arr.get(i)) - 'a'] == 1) return arr.get(i);  
        }
        
        return -1;
    }
}

Solution

遍历，建立哈希表，记录出现次数。
再次遍历，如果出现次数为1，则返回下标。

Code 2

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class Solution {
    public int firstUniqChar(String s) {
        HashMap<Character,Integer> map = new HashMap<Character,Integer>();
        for ( int i = 0; i < s.length(); i++ ){
            char curChar = s.charAt(i);
            if ( !map.containsKey(curChar) ){
                map.put(curChar, 1);
            }
            else{
                map.put(curChar, map.get(curChar)+1);
            }
        }
        
        for ( int i = 0; i < s.length(); i++ ){
            char curChar = s.charAt(i);
            if ( map.get(curChar) == 1 ){
                return i;
            }
        }
        return -1;
    }
}
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