LeetCode 387---First Unique Character in a String

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

Example 1:

Input: s = “leetcode”
Output: 0

Example 2:

Input: s = “loveleetcode”
Output: 2

Example 3:

Input: s = “aabb”
Output: -1

Constraints:

• 1 <= s.length <= 105
• s onsists of only lowercase English letters.

Solution1 using java API indexOf

class Solution {
    public int firstUniqChar(String s) {
        for (int i = 0; i < s.length(); i++) {
            if (s.indexOf(s.charAt(i)) == s.lastIndexOf(s.charAt(i))) {
                return i;
            }
        }
        return -1;
    }
}
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Solution2 using HashMap

class Solution {
    public int firstUniqChar(String s) {
        int res = -1;
        HashMap<Character, Integer> map1 = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            if (map1.get(s.charAt(i)) == null) {
                map1.put(s.charAt(i), 1);
            } else {
                map1.put(s.charAt(i), map1.get(s.charAt(i)) + 1);
            }
        }
        for (int j = 0; j < s.length(); j++) {
            if (map1.get(s.charAt(j)) == 1) {
                res = j;
                break;
            }
        }
        return res;
    }
}
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Solution3 dual circulation

class Solution {
    public int firstUniqChar(String s) {
        int res = -1;
        boolean isDouble = false;
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j < s.length(); j++) {
                if (i == j) {
                    continue;
                }
                if (s.charAt(i) == s.charAt(j)) {
                    isDouble = true;
                    break;
                }
            }
            if (isDouble) {
                isDouble = false;
            } else {
                res = i;
                break;
            }
        }
        return res;
    }
}
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