Given an integer array `nums`

, return the length of the longest strictly increasing subsequence.

A **subsequence** is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, `[3,6,2,7]`

is a subsequence of the array `[0,3,1,6,2,2,7]`

.

**Example 1:**

Input:nums = [10,9,2,5,3,7,101,18]Output:4Explanation:The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

**Example 2:**

Input:nums = [0,1,0,3,2,3]Output:4

**Example 3:**

Input:nums = [7,7,7,7,7,7,7]Output:1

**Constraints:**

`1 <= nums.length <= 2500`

`-10`

^{4}<= nums[i] <= 10^{4}

**Follow up:**

- Could you come up with the
`O(n`

solution?^{2}) - Could you improve it to
`O(n log(n))`

time complexity?

```
struct Solution;
impl Solution {
fn length_of_lis(nums: Vec<i32>) -> i32 {
let mut dp: Vec<i32> = vec![];
for x in nums {
if let Err(i) = dp.binary_search(&x) {
if i == dp.len() {
dp.push(x)
} else {
dp[i] = x;
}
}
}
dp.len() as i32
}
}
#[test]
fn test() {
let nums = vec![10, 9, 2, 5, 3, 7, 101, 18];
let res = 4;
assert_eq!(Solution::length_of_lis(nums), res);
}
```