Go Java算法之字符串中第一个唯一字符详解

class Solution {
    public int firstUniqChar(String s) {
        Map<Character, Integer> frequency = new HashMap<Character, Integer>();
        for (int i = 0; i < s.length(); ++i) {
            char ch = s.charAt(i);
            frequency.put(ch, frequency.getOrDefault(ch, 0) + 1);
        }
        for (int i = 0; i < s.length(); ++i) {
            if (frequency.get(s.charAt(i)) == 1) {
                return i;
            }
        }
        return -1;
    }
}

type pair struct {
    ch  byte
    pos int
}
func firstUniqChar(s string) int {
    n := len(s)
    pos := [26]int{}
    for i := range pos[:] {
        pos[i] = n
    }
    q := []pair{}
    for i := range s {
        ch := s[i] - 'a'
        if pos[ch] == n {
            pos[ch] = i
            q = append(q, pair{ch, i})
        } else {
            pos[ch] = n + 1
            for len(q) > 0 && pos[q[0].ch] == n+1 {
                q = q[1:]
            }
        }
    }
    if len(q) > 0 {
        return q[0].pos
    }
    return -1
}