[jobdu]第一个只出现一次的字符

刚开始还在想双指针,完全没必要。然后最土的可以扫两遍O(n),用一个数组记录出现次数。然后想到如果数组里记录出现的位置,可以只扫一遍O(n),第二遍扫字符集就行了。

#include <iostream>
#include <string>
using namespace std;
 
#define INF 1<<30
int main()
{
    string s;
    while (cin >> s)
    {
        int len = s.length();
        if (len == 0)
        {
            cout << "-1" << endl;
            continue;
        }
        int count[26];
        for (int i = 0; i < 26; i++)
        {
            count[i] = -1;
        }
        for (int i = 0; i < len; i++)
        {
            if (count[s[i]-'A'] == -1)
            {
                count[s[i]-'A'] = i; // save the first position
            }
            else
            {
                count[s[i]-'A'] = INF;
            }
        }
        int min = INF;
        for (int i = 0; i < 26; i++)
        {
            if (min > count[i] && count[i] != -1) min = count[i];
        }
        if (min != INF) cout << min << endl;
        else cout << -1 << endl;
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/lautsie/p/3404468.html