2022-05-12：小歪每次会给你两个字符串：
笔记s1和关键词s2，请你写一个函数，
判断s2的排列之一是否是s1的子串。
如果是，返回true；
否则，返回false。
来自字节飞书团队。
答案2022-05-12：
欠债表。滑动窗口，str2存map。
代码用rust编写。代码如下：
代码语言：rust复制fn main() {
let ans=check2("fabkbbca","abcb");
println!("ans = {}",ans);
}
fn check2( s1:&str, s2:&str) ->bool{
if s1.chars().count() < s2.chars().count() {
return false;
}
let str2: Vec<char> = s2.chars().collect::<Vec<_>>();
let mut count:[isize;256]=[0;256];
for i in 0..s2.chars().count() {
count[str2[i] as usize] =1;
}
let m = s2.chars().count() as isize;
let st1: Vec<char> = s1.chars().collect::<Vec<_>>();
let mut in_valid_times:isize = 0;
let mut r:isize = 0;
while r < m {
if count[st1[r as usize] as usize] <= 0 {
in_valid_times =1;
}
count[st1[r as usize] as usize]-=1;
r =1;
}
while r < st1.len() as isize {
if in_valid_times == 0 {
return true;
}
if count[st1[r as usize] as usize] <= 0 {
in_valid_times =1;
}
count[st1[r as usize] as usize]-=1;
if count[st1[(r - m) as usize] as usize] < 0 {
in_valid_times-=1;
}
count[st1[(r - m) as usize] as usize] =1;
r =1;
}
return in_valid_times == 0;
}执行结果如下：
在这里插入图片描述左神java代码
