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Question

Formatted question description: https://leetcode.ca/all/387.html

 387	First Unique Character in a String

 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.

Algorithm

The hash table builds a mapping between each character and its number of occurrences, and then traverses the string in order, finds the first character with a number of occurrences of 1, and returns its position.

Code

• Java
• C++
• Python

• 
  public class First_Unique_Character_in_a_String {
  
      public static void main (String[] args) {
          First_Unique_Character_in_a_String out = new First_Unique_Character_in_a_String();
          Solution s = out.new Solution();
  
          System.out.println(s.firstUniqChar("loveleetcode"));
      }
  
      class Solution {
          public int firstUniqChar(String s) {
  
              if (s == null || s.length() == 0) {
                  return -1;
              }
  
              // unique[i] meaning index i has no duplicate
              int[] count = new int[256];
  
              for (char each: s.toCharArray()) {
                  count[each] += 1;
              }
  
              for (int i = 0; i < s.length(); i++) {
                  if (count[s.charAt(i)] == 1) {
                      return i;
                  }
              }
  
              return -1;
          }
      }
  }
  

• // OJ: https://leetcode.com/problems/first-unique-character-in-a-string/
  // Time: O(S)
  // Space: O(1)
  class Solution {
  public:
      int firstUniqChar(string s) {
          int cnt[26] = {0};
          for (char c : s) cnt[c - 'a']++;
          for (int i = 0; i < s.size(); ++i) {
              if (cnt[s[i] - 'a'] == 1) return i;
          }
          return -1;
      }
  };
  

• class Solution:
      def firstUniqChar(self, s: str) -> int:
          counter = Counter(s)
          for i, c in enumerate(s):
              if counter[c] == 1:
                  return i
          return -1
  
  ############
  
  class Solution(object):
    def firstUniqChar(self, s):
      """
      :type s: str
      :rtype: int
      """
      letters = [0] * 26
      for c in s:
        ci = ord(c) - ord('a')
        letters[ci] += 1
      for i in range(0, len(s)):
        ci = ord(s[i]) - ord('a')
        if letters[ci] == 1:
          return i
      return -1
  
  

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