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。
Java: Brute Force
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:
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:
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:
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:
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:
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
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
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
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])
C++:
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; } };