## 673. Number of Longest Increasing Subsequence

Given an integer array `nums`, return the number of longest increasing subsequences.

Notice that the sequence has to be strictly increasing.

Example 1:

```Input: nums = [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
```

Example 2:

```Input: nums = [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.

```

Constraints:

• `1 <= nums.length <= 2000`
• `-106 <= nums[i] <= 106`

## Rust Solution

``````struct Solution;

impl Solution {
fn find_number_of_lis(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut len: Vec<usize> = vec![1; n];
let mut cnt: Vec<usize> = vec![1; n];
for i in 0..n {
for j in 0..i {
if nums[j] < nums[i] {
if len[j] + 1 == len[i] {
cnt[i] += cnt[j];
}
if len[j] == len[i] {
len[i] += 1;
cnt[i] = cnt[j];
}
}
}
}
let max_len = *len.iter().max().unwrap();
let mut res = 0;
for i in 0..n {
if len[i] == max_len {
res += cnt[i];
}
}
res as i32
}
}

#[test]
fn test() {
let nums = vec![1, 3, 5, 4, 7];
let res = 2;
assert_eq!(Solution::find_number_of_lis(nums), res);
let nums = vec![2, 2, 2, 2, 2];
let res = 5;
assert_eq!(Solution::find_number_of_lis(nums), res);
}
``````

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