hihocoder offer收割编程练习赛11 A hiho字符串

思路：

我用的尺取。

注意题目描述为恰好2个'h'，1个'i'，1个'o'。

实现：

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;

int ch[100005], ci[100005], co[100005];

bool check(int pos, int a, int b, int c)
{
    return ch[pos] <= a - 2 && ci[pos] <= b - 1 && co[pos] <= c - 1;
}

int main()
{
    string s;
    cin >> s;
    int n = s.length();
    ch[0] += (s[0] == 'h');
    ci[0] += (s[0] == 'i');
    co[0] += (s[0] == 'o');
    int last = 0, min_len = INF;
    for (int i = 1; i < n; i++)
    {
        ch[i] = ch[i - 1] + (s[i] == 'h');
        ci[i] = ci[i - 1] + (s[i] == 'i');
        co[i] = co[i - 1] + (s[i] == 'o');
        if (ch[i] >= 2 && ci[i] >= 1 && co[i] >= 1)
        {
            while (last < i && check(last, ch[i], ci[i], co[i]))
            {
                last++;
            }
            if (ch[i] - ch[last - 1] == 2 && 
                ci[i] - ci[last - 1] == 1 && 
                co[i] - co[last - 1] == 1) 
                min_len = min(min_len, i - last + 1);
        }
    }
    if (min_len != INF)
        cout << min_len << endl;
    else
        cout << -1 << endl;
    return 0;
}