05 August 2008
Given a string, find the first non-repeating character in it and return it’s index. If it doesn’t exist, return -1.
Examples:
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s = "leetcode"
return 0.
s = "loveleetcode",
return 2.
Note: You may assume the string contain only lowercase letters.
找到全小写字符串中第一个出现的非重复字符，那么就先找到每个字符出现的次数，然后找到出现次数为1的字符，返回其位置即可。由于Java中的HashMap的顺序是按照有利于随机查找的hash的顺序，并非是按照输入的顺序，所以用一个26长度的数组来记录频率。
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class Solution {
public int firstUniqChar(String s) {
if (s == null || s.length() == 0) {
return -1;
}
int[] freq = new int[26];
for (char ch : s.toCharArray()) {
freq[ch - 'a']++;
}
for (int i = 0; i < s.length(); i++) {//注意依然按照s中字符的顺序来遍历
if (freq[s.charAt(i) - 'a'] == 1) return i;
}
return -1;
}
}