2022-07-21：给定一个字符串str，和一个正数k，
你可以随意的划分str成多个子串，
目的是找到在某一种划分方案中，有尽可能多的回文子串，长度>=k，并且没有重合。
返回有几个回文子串。
来自optiver。
答案2022-07-21：
马拉车算法 贪心。
代码用rust编写。代码如下：
代码语言：rust复制use rand::Rng;
fn main() {
let n: i32 = 20;
let r = 3;
let test_time: i32 = 50000;
println!("测试开始");
for i in 0..test_time {
let str = random_string(n, r);
let k = rand::thread_rng().gen_range(0, str.len() as i32) 1;
let ans1 = max1(&str, k);
let ans2 = max2(&str, k);
if ans1 != ans2 {
println!("i = {}", i);
println!("str = {}", str);
println!("k = {}", k);
println!("ans1 = {}", ans1);
println!("ans2 = {}", ans2);
println!("出错了!");
break;
}
}
println!("测试结束");
}
// 暴力尝试
// 为了测试
// 可以改成动态规划，但不是最优解
fn max1(s: &str, k: i32) -> i32 {
if s.len() == 0 {
return 0;
}
let mut str = s.as_bytes().to_vec();
return process1(&mut str, 0, k);
}
fn process1(str: &mut Vec<u8>, index: i32, k: i32) -> i32 {
if str.len() as i32 - index < k {
return 0;
}
let mut ans = process1(str, index 1, k);
for i in index k - 1..str.len() as i32 {
if is_palindrome(str, index, i) {
ans = get_max(ans, 1 process1(str, i 1, k));
}
}
return ans;
}
fn is_palindrome(str: &mut Vec<u8>, mut ll: i32, mut rr: i32) -> bool {
while ll < rr {
if str[ll as usize] != str[rr as usize] {
return false;
}
ll = 1;
rr -= 1;
}
return true;
}
fn get_max<T: Clone Copy std::cmp::PartialOrd>(a: T, b: T) -> T {
if a > b {
a
} else {
b
}
}
fn get_min<T: Clone Copy std::cmp::PartialOrd>(a: T, b: T) -> T {
if a < b {
a
} else {
b
}
}
// 最优解
// 时间复杂度O(N)
fn max2(s: &str, k: i32) -> i32 {
if s.len() == 0 {
return 0;
}
let mut str = manacher_string(s);
let mut p = vec![];
for _ in 0..str.len() as i32 {
p.push(0);
}
let mut ans = 0;
let mut next = 0;
// k == 5 回文串长度要 >= 5
// next == 0
// 0.... 8 第一块！
// next -> 9
// 9.....17 第二块！
// next -> 18
// 18....23 第三块
// next一直到最后!
next = manacher_find(&mut str, &mut p, next, k);
while next != -1 {
next = if str[next as usize] == ('#' as u8) {
next
} else {
next 1
};
ans = 1;
next = manacher_find(&mut str, &mut p, next, k);
}
return ans;
}
fn manacher_string(s: &str) -> Vec<u8> {
let str = s.as_bytes().to_vec();
let mut ans: Vec<u8> = vec![];
for _ in 0..str.len() as i32 * 2 1 {
ans.push(0);
}
let mut index: i32 = 0;
for i in 0..ans.len() as i32 {
if (i & 1) == 0 {
ans[i as usize] = '#' as u8;
} else {
ans[i as usize] = str[index as usize];
index = 1;
}
}
return ans;
}
// s[l...]字符串只在这个范围上，且s[l]一定是'#'
// 从下标l开始，之前都不算，一旦有某个中心回文半径>k，马上返回右边界
fn manacher_find(s: &mut Vec<u8>, p: &mut Vec<i32>, l: i32, k: i32) -> i32 {
let mut c = l - 1;
let mut r = l - 1;
let n = s.len() as i32;
for i in l..s.len() as i32 {
p[i as usize] = if r > i {
get_min(p[(2 * c - i) as usize], r - i)
} else {
1
};
while i p[i as usize] < n
&& i - p[i as usize] > l - 1
&& s[(i p[i as usize]) as usize] == s[(i - p[i as usize]) as usize]
{
p[i as usize] = 1;
if p[i as usize] > k {
return i k;
}
}
if i p[i as usize] > r {
r = i p[i as usize];
c = i;
}
}
return -1;
}
// 为了测试
fn random_string(n: i32, r: i32) -> String {
let mut ans: String = String::from("");
let ans_len = rand::thread_rng().gen_range(1, n);
for _ in 0..ans_len {
ans.push((rand::thread_rng().gen_range(0, r) 'a' as i32) as u8 as char);
}
return ans;
}执行结果如下：
在这里插入图片描述左神java代码
