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] <= 106struct 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);
}