LeetCode387 First Unique Character in a String

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题目

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.
即找到字符串中第一个不重复的字母。

方法一

很自然的想到两次遍历来找到唯一的字符，同时利用字母只有26个的特点减少循环次数。

public int firstUniqChar(String s) {        if (s.length() == 0) return -1;        int[] test = new int[s.length()];        int count = 0;        int result = -1;        boolean found = true;        for (int i = 0; i < s.length() && count < 27; i++) {            if (test[i] == 1) continue;            else count++;            for (int j = i + 1; j < s.length(); j++) {                if (test[j] == 1) continue;                if (s.charAt(i) == s.charAt(j)) {                    found = false;                    test[i] = 1;                    test[j] = 1;                }            }            if (found) {                result = i;                break;            }            found = true;        }        return result;    }

方法二

嵌套循环毫无疑问效率低下，这是想到使用键值对记录每个字母的出现次数，同时利用hashmap去重的特性。

public int firstUniqChar1(String s) {        if (s.length() == 0) return -1;        int result = -1;        Map<Character, Integer> map = new HashMap<>();        for (int i = 0; i < s.length(); i++) {            char c = s.charAt(i);            if (map.containsKey(c))                map.put(c, map.get(c) + 1);            else map.put(c, 1);        }        for (int i = 0; i < s.length(); i++) {            char c = s.charAt(i);            if (map.get(c) == 1) {                result = i;                break;            }        }        return result;    }

虽然简化了步骤，但是map的使用仍然需要较多的时间开销。这是可以使用数组的1~26个位置代表a~z出现的次数。

public int firstUniqChar2(String s) {        if (s.length() == 0) return -1;        int result = -1;        int[] record = new int[26];        for (int i = 0; i < s.length(); i++) {            char c = s.charAt(i);            int index = c - 'a';            record[index]++;        }        for (int i = 0; i < s.length(); i++) {            char c = s.charAt(i);            if (record[s.charAt(i)-'a'] == 1) {                result = i;                break;            }        }        return result;    }

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