2022-10-01：给定一个字符串 s，计算 s 的 不同非空子序列 的个数 因为结果可能很大，所以返回答案需要对 10^9 + 7 取余 。 字符串的 子序列 是经由原字符串删除一些（也可能不删除

use std::collections::HashMap;fn main() {    let ans = distinct_subseq_ii("bccaccbaabbc");    println!("ans = {}", ans);    let ans = zuo("bccaccbaabbc");    println!("ans = {}", ans);}
fn distinct_subseq_ii(s: &str) -> i32 {    if s.len() == 0 {        return 0;    }    let m = 1000000007;    let str2: Vec<u8> = s.bytes().collect();    // a -> 0 ->    // b -> 1 ->    // c -> 2 ->    // z -> 25 ->    // count[i] = 0    let mut count = [0; 26];    // { }    let mut all = 1; // 算空集    for x in str2.iter() {        // x        // 纯新增        // add = all - count[x]        // all += add        // count[x] + add        let add = (all - count[(*x - 'a' as u8) as usize] + m) % m;        all = (all + add) % m;        count[(*x - 'a' as u8) as usize] = (count[(*x - 'a' as u8) as usize] + add) % m;    }    // {} 去掉！    return all - 1;}fn zuo(s: &str) -> i32 {    if s.len() == 0 {        return 0;    }    let m = 1000000007;    let str2: Vec<u8> = s.bytes().collect();    let mut map: HashMap<u8, i32> = HashMap::new();    let mut all = 1; // 一个字符也没遍历的时候，有空集    for x in str2.iter() {        let new_add = all;        //      int curAll = all + newAdd - (map.containsKey(x) ? map.get(x) : 0);        let mut cur_all = all;        cur_all = (cur_all + new_add) % m;        cur_all = (cur_all            - if map.contains_key(x) {                *map.get(x).unwrap()            } else {                0            }            + m)            % m;        all = cur_all;        map.insert(*x, new_add);    }    return all - 1;}