[LintCode/LeetCode] First Unique Character in a S

Xufc 发布于2019-08-15 13:27 / 3265人阅读

Problem

Given a string, find the first non-repeating character in it and return it"s index. If it doesn"t exist, return -1.

Example

Given s = "lintcode", return 0.

Given s = "lovelintcode", return 2.

Tags

Amazon Microsoft Bloomberg

Solution

a fast way

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

a dumb way...

public class Solution {
    public int firstUniqChar(String s) {
        //store string in an array
        char[] str = s.toCharArray();
        //use HashMap to check each character"s frequency
        Map map = new HashMap<>();
        for (int i = 0; i < str.length; i++) {
            char ch = str[i];
            //reset duplicate chars to "#"
            if (map.containsKey(ch)) {
                str[map.get(ch)] = "#";
                str[i] = "#";
            } else {
                map.put(ch, i);
            }
        }
        int index = str.length;
        for (int i = 0; i < str.length; i++) {
            //find the first character that is not "#" and return its index
            if (str[i] != "#") {
                return i;
            }
        }
        //if no unique character, return -1
        return -1;
    }
}

混合云云服务器 array_unique character unique numeric-character-re

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