387. First Unique Character in a String

hashmap O(n)
这个方法比简单统计字母频要快

class Solution {
public:
    int firstUniqChar(string s) {
        int f[26] = {0};
        for (const auto &c : s) {
            ++f[c - 'a'];
        }
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            if (f[s[i] - 'a'] == 1) {
                return i;
            }
        }
        return -1;
    }
};

把重复的字母的下标全部记成n，不重复的则记成i，然后遍历找出最小的i

class Solution {
public:
    int firstUniqChar(string s) {
        vector<int> m(26, INT_MAX);
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            int k = s[i] - 'a';
            if (m[k] == INT_MAX) {
                m[k] = i;
            } else if (m[k] < n) {
                m[k] = n;
            }
        }
        int res = *min_element(m.begin(), m.end());
        return res < n ? res : -1;
    }
};