LeetCode Top Interview Questions 387. First Unique Character in a String (Java版; Easy)

welcome to my blog

LeetCode Top Interview Questions 387. First Unique Character in a String (Java版; Easy)

题目描述

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.
Note: You may assume the string contain only lowercase letters.

第一次做;bit-map记录每个字符出现的次数, 位置信息在原始字符串中; 下面的那个做法在处理位置信息时愚蠢了…

class Solution {
    public int firstUniqChar(String s) {
        if(s==null||s.length()==0)
            return -1;
        //
        int[] arr = new int[26];
        for(int i=0; i<s.length(); i++){
            char ch = s.charAt(i);
            arr[ch-'a']++;
        }
        for(int i=0; i<s.length(); i++){
            char ch = s.charAt(i);
            if(arr[ch-'a']==1)
                return i;
        }
        return -1;
    }
}

第一次做;bit-map记录出现次数; StringBuilder记录位置信息(这么做就麻烦了… 直接再遍历一次原始字符串即可); 时间复杂度O(N), 空间复杂度O(N)

/**
26个英文字母, 使用bit-map, 记录每个字符出现的次数, 再配合一个StringBuilder记录扫描过的字符从而保留位置信息
向面试官确认一下是否只有26个英文小写/大写字母
如果是全体ascii的话就需要一个更大一些的数组
*/
class Solution {
    public int firstUniqChar(String s) {
        if(s==null||s.length()==0)
            return -1;
        //
        int[] arr = new int[26];
        StringBuilder sb = new StringBuilder();
        for(int i=0; i<s.length(); i++){
            char ch = s.charAt(i);
            arr[ch-'a']++;
            sb.append(ch);
        }
        for(int i=0; i<sb.length(); i++){
            char ch = sb.charAt(i);
            if(arr[ch-'a']==1)
                return i;
        }
        return -1;
        
    }
}

leetcode题解中关于该问题的其它思考

Good O(n) solution, but it still can be improved, since you read through the whole array twice. 
Take an example of DNA sequence: there could be millions of letters long with just 4 alphabet 
letter. What happened if the non-repeated letter is at the end of DNA sequence? This would 
dramatically increase your running time since we need to scan it again.

可以使用两个哈希表解决上述问题, 一个哈希表记录出现次数, 另一个哈希表记录索引. 最后遍历第一个哈希表; 我对遍历哈希表的语法不熟

class Solution {
    public int firstUniqChar(String s) {
        Map<Character, Integer> feq = new HashMap<>();
        Map<Character, Integer> index = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            feq.put(s.charAt(i), feq.getOrDefault(s.charAt(i), 0) + 1);
            index.put(s.charAt(i), i);
        }
        
        int min = Integer.MAX_VALUE;     
        for (char c : feq.keySet()) {
            if (feq.get(c) == 1) {
                min = Math.min(min, index.get(c));
            }
        }
        
        return min == Integer.MAX_VALUE ? -1 : min;
    }
}