2022-07-07：原本数组中都是大于0、小于等于k的数字，是一个单调不减的数组，
其中可能有相等的数字，总体趋势是递增的。
但是其中有些位置的数被替换成了0，我们需要求出所有的把0替换的方案数量：
1）填充的每一个数可以大于等于前一个数，小于等于后一个数；
2）填充的每一个数不能大于k。
来自腾讯音乐。
答案2022-07-07：
方法一：动态规划。
方法二：数学方法。用到组合，C(b-a m,m)。
代码用rust编写。代码如下：
代码语言：javascript复制use rand::Rng;
fn main() {
let nn: i64 = 20;
let kk: i64 = 30;
let test_time: i32 = 10000;
println!("测试开始");
for i in 0..test_time {
let n = rand::thread_rng().gen_range(0, nn) 1;
let k = rand::thread_rng().gen_range(0, kk) 1;
let mut arr = random_array(n, k);
let ans1 = ways1(&mut arr, k);
let ans2 = ways2(&mut arr, k);
if ans1 != ans2 {
println!("出错了!{}", i);
println!("ans1 = {}", ans1);
println!("ans2 = {}", ans2);
break;
}
}
println!("测试结束");
}
// 动态规划
fn ways1(nums: &mut Vec<i64>, k: i64) -> i64 {
let n = nums.len() as i64;
// dp[i][j] : 一共i个格子，随意填，但是不能降序，j种数可以选
let mut dp: Vec<Vec<i64>> = vec![];
for i in 0..n 1 {
dp.push(vec![]);
for _ in 0..k 1 {
dp[i as usize].push(0);
}
}
for i in 1..=n {
dp[i as usize][1] = 1;
}
for i in 1..=k {
dp[1][i as usize] = i;
}
for i in 2..=n {
for j in 2..=k {
dp[i as usize][j as usize] =
dp[(i - 1) as usize][j as usize] dp[i as usize][(j - 1) as usize];
}
}
let mut res = 1;
let mut i: i64 = 0;
let mut j: i64 = 0;
while i < nums.len() as i64 {
if nums[i as usize] == 0 {
j = i 1;
while j < nums.len() as i64 && nums[j as usize] == 0 {
j = 1;
}
let left_value = if i - 1 >= 0 {
nums[(i - 1) as usize]
} else {
1
};
let right_value = if j < nums.len() as i64 {
nums[j as usize]
} else {
k
};
res *= dp[(j - i) as usize][(right_value - left_value 1) as usize];
i = j;
}
i = 1;
}
return res;
}
// 数学方法
// a ~ b范围的数字随便选，可以选重复的数，一共选m个
// 选出有序序列的方案数：C ( m, b - a m )
fn ways2(nums: &mut Vec<i64>, k: i64) -> i64 {
let mut res = 1;
let mut i: i64 = 0;
let mut j: i64 = 0;
while i < nums.len() as i64 {
if nums[i as usize] == 0 {
j = i 1;
while j < nums.len() as i64 && nums[j as usize] == 0 {
j = 1;
}
let left_value = if i - 1 >= 0 {
nums[(i - 1) as usize]
} else {
1
};
let right_value = if j < nums.len() as i64 {
nums[j as usize]
} else {
k
};
let numbers = j - i;
res *= c(right_value - left_value numbers, numbers);
i = j;
}
i = 1;
}
return res;
}
// 从一共a个数里，选b个数，方法数是多少
fn c(a: i64, b: i64) -> i64 {
if a == b {
return 1;
}
let mut x = 1;
let mut y = 1;
let mut i = b 1;
let mut j = 1;
while i <= a {
x *= i;
y *= j;
let gcd = gcd(x, y);
x /= gcd;
y /= gcd;
i = 1;
j = 1;
}
return x / y;
}
fn gcd(m: i64, n: i64) -> i64 {
if n == 0 {
m
} else {
gcd(n, m % n)
}
}
// 为了测试
fn random_array(n: i64, k: i64) -> Vec<i64> {
let mut ans: Vec<i64> = vec![];
for _i in 0..n {
ans.push(rand::thread_rng().gen_range(0, k) 1);
}
ans.sort();
for i in 0..n {
ans[i as usize] = if rand::thread_rng().gen_range(0, 2) == 0 {
0
} else {
ans[i as usize]
};
}
return ans;
}执行结果如下：
***
[左神java代码](https://github.com/algorithmzuo/weekly-problems/blob/main/src/class_2022_04_3_week/Code03_ValidSortedArrayWays.java)
