LeetCode（剑指 Offer）- 50. 第一个只出现一次的字符

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题目大意：略

解题思路：略

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AC 代码

• Java

// 解决方案(1)
class Solution {
    public char firstUniqChar(String s) {
        HashMap<Character, Boolean> dic = new HashMap<>();
        char[] sc = s.toCharArray();
        for(char c : sc)
            dic.put(c, !dic.containsKey(c));
        for(char c : sc)
            if(dic.get(c)) return c;
        return ' ';
    }
}

// 解决方案(2)
class Solution {
    public char firstUniqChar(String s) {
        Map<Character, Boolean> dic = new LinkedHashMap<>();
        char[] sc = s.toCharArray();
        for(char c : sc)
            dic.put(c, !dic.containsKey(c));
        for(Map.Entry<Character, Boolean> d : dic.entrySet()){
           if(d.getValue()) return d.getKey();
        }
        return ' ';
    }
}

// 解决方案(3)
class Solution {
    public char firstUniqChar(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            cnt[c - 'a']++;
        }
        for (int i = 0; i < s.length(); i++) {
            if (cnt[s.charAt(i) - 'a'] == 1) {
                return s.charAt(i);
            }
        }
        return ' ';
    }
}

• C++

// 解决方案(1)
class Solution {
public:
    char firstUniqChar(string s) {
        unordered_map<char, bool> dic;
        for(char c : s)
            dic[c] = dic.find(c) == dic.end();
        for(char c : s)
            if(dic[c]) return c;
        return ' ';
    }
};

// 解决方案(2)
class Solution {
public:
    char firstUniqChar(string s) {
        vector<char> keys;
        unordered_map<char, bool> dic;
        for(char c : s) {
            if(dic.find(c) == dic.end())
                keys.push_back(c);
            dic[c] = dic.find(c) == dic.end();
        }
        for(char c : keys) {
            if(dic[c]) return c;
        }
        return ' ';
    }
};