2022-09-19：给定字符串 S and T，找出 S 中最短的（连续）子串 W ，使得 T 是 W 的 子序列 。 如果 S 中没有窗口可以包含 T 中的所有字符，返回空字符串 ““。 如果有不

fn main() {    let s = "xxaxxbxxcxxaxbyc";    let t = "abc";    let ans = min_window4(s, t);    println!("ans = {}", ans);}
const MAX_VALUE: i32 = 1 << 31 - 1;
fn min_window4(s: &str, t: &str) -> String {    let str = s.as_bytes();    let target = t.as_bytes();    let n = str.len() as i32;    let m = target.len() as i32;    let mut dp: Vec<Vec<i32>> = vec![];    for i in 0..n + 1 {        dp.push(vec![]);        for _ in 0..m + 1 {            dp[i as usize].push(0);        }    }    for si in 0..=n {        dp[si as usize][m as usize] = si;    }    for ti in 0..m {        dp[n as usize][ti as usize] = MAX_VALUE;    }    let mut si = n - 1;    while si >= 0 {        let mut ti = m - 1;        while ti >= 0 {            let r1 = dp[(si + 1) as usize][ti as usize];            let r2 = if str[si as usize] == target[ti as usize] {                dp[(si + 1) as usize][(ti + 1) as usize]            } else {                MAX_VALUE            };            dp[si as usize][ti as usize] = get_min(r1, r2);            ti -= 1;        }        si -= 1;    }    let mut len = MAX_VALUE;    let mut l = -1;    let mut r = -1;    for si in 0..str.len() as i32 {        let right = dp[si as usize][0];        if right != MAX_VALUE && right - si < len {            len = dp[si as usize][0] - si;            l = si;            r = right;        }    }    return if l == -1 {        String::from("")    } else {        String::from(&s[l as usize..r as usize])    };}fn get_min<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T {    if a < b {        a    } else {        b    }}