one

#include <bits/stdc++.h>

using namespace std;

int main() {
    int judge[270];
    int num = 0;
    int acc = 0;
    int l = 0;
    int r = 0;
    string c;
    cin >> c;
    int len = c.size();
    int result = len + 1;
    memset(judge, 0, sizeof(judge));
    while (l < c.size()) {
        while (r < c.size() && acc < 26) {
            judge[c[r] - 'a']++;
            if (judge[c[r++] - 'a'] == 1) {
                acc++;
            }
        }
        if (acc == 26) {
            result = min(result, r - l + 1);
            if (judge[c[l] - 'a'] == 1) acc--;
        }
        judge[c[l] - 'a']--;
        l++;
    }
    cout << result;
    return 0;
}

c++stl库中string.length()和string.size()的坑

length()和size()的返回值都是size_t类型，也就是unsigned int，在遇到下面情况，会出现不符合你预想的情况：

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

int main(){
	string haystack = "";
	string needdle = "a";
	int i=0;
	cout<<haystack.size()<<endl;
	cout<<needdle.size()<<endl;
	cout<<haystack.size()-needdle.size();
	return 0;
}

//输出为false

*signed int和unsigned int运算时，会进行类型提升，*signed int会提升为unsigned int，-1就变成了4294967295。

另外，string的length和size方法没有任何区别。 ———————————————— 版权声明：本文为CSDN博主「乐观的神」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。 原文链接：https://blog.csdn.net/qq_35926606/article/details/116503944

two

链接