[LeetCode] 387. First Unique Character in a String 字符串的第一个唯一字符

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.

给一个字符串，找出第一个不重复的字符，返回它的index，如果不存在，返回-1。假设字符都是小写字母。

解法1: 暴力搜索， T：O(n^2)

解法2: HashMap，第一次遍历每一个字符，统计每种字符出现的次数。第二次遍历，找到第一出现的字符次数为1的字符。

解法3: int [26]数组，由于只有26个字母，所以可以用数组来统计出现的个数。

解法4: 循环26个字母，统计每个字母出现次数为1的字母，写入按出现的index排序的数组。然后取数组里最前面的那个或者为空时是-1

Java: Brute Force

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class Solution {     public static int firstUniqChar(String s){         for (int i = 0; i < s.length(); i++) {             boolean isUnique = true;             for (int j = 0; j < s.length(); j++) {                 if (i != j && s.charAt(i) == s.charAt(j)){                     isUnique = false;                     break;                 }             }             if (isUnique) return i;         }         return -1;     } }

Java:

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class Solution {     public int firstUniqChar(String s){         Map<Character, Integer> charMap = new HashMap<>(s.length()); //预先分配大小,避免扩容性能影响         for (int i = 0; i < s.length(); i++) {             if (!charMap.containsKey(s.charAt(i))){                 charMap.put(s.charAt(i), 1);             }else {                 charMap.put(s.charAt(i), charMap.get(s.charAt(i))+1);             }         }           for (int i = 0; i < s.length(); i++) {             if (charMap.get(s.charAt(i)) == 1){                 return i;             }         }         return -1;     } }

Java:

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public class Solution {      public int firstUniqChar(String s) {          char[] array = s.toCharArray();          int[] a = new int[26];          for(int i=0;i<s.length();i++)a[array[i]-'a']++;          for(int i=0;i<s.length();i++){              if(a[array[i]-'a']==1){                  return i;              }          }          return -1;      }  }

Java:

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class Solution {     public int firstUniqChar(string s) {          vector<int> count(26);          for(int i=0;i<s.size();i++)              count[s[i]-'a']++;          for(int i=0;i<s.size();i++)              if(count[s[i]-'a']==1)                  return i;          return -1;      }  }

Java:　　

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class Solution {     public int firstUniqChar(String s) {        for(int i = 0; i<s.length(); i++) {          if(s.lastIndexOf(s.charAt(i))==s.indexOf(s.charAt(i))) return i;         }         return -1;      }  }

Python:

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class Solution(object):     def firstUniqChar(self, s):         """         :type s: str         :rtype: int         """         letters = {}         for c in s:             if c in letters:                 letters[c] = letters[c] + 1             else:                 letters[c] = 1            for i in xrange(len(s)):             if letters[s[i]] == 1:                 return i         return -1

Python: 162ms

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from collections import defaultdict   class Solution(object):     def firstUniqChar(self, s):         """         :type s: str         :rtype: int         """         lookup = defaultdict(int)         candidtates = set()         for i, c in enumerate(s):             if lookup[c]:                 candidtates.discard(lookup[c])             else:                 lookup[c] = i+1                 candidtates.add(i+1)           return min(candidtates)-1 if candidtates else -1

Python: 92ms

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class Solution(object):     def firstUniqChar(self, s):         """         :type s: str         :rtype: int         """         return min([s.find(c) for c in 'abcdefghijklmnopqrstuvwxyz' if s.count(c)==1] or [-1])

Python: 75ms

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class Solution(object):     def firstUniqChar(self, s):         """         :type s: str         :rtype: int         """         return min([s.find(c) for c in string.ascii_lowercase if s.count(c)==1] or [-1])

Python: 60ms

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def firstUniqChar(self, s):         """         :type s: str         :rtype: int         """                   letters='abcdefghijklmnopqrstuvwxyz'         index=[s.index(l) for l in letters if s.count(l) == 1]         return min(index) if len(index) > 0 else -1

C++:

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class Solution { public:     int firstUniqChar(string s) {         unordered_map<char, int> m;         for (char c : s) ++m[c];         for (int i = 0; i < s.size(); ++i) {             if (m[s[i]] == 1) return i;         }         return -1;     } };

　　

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