## 1143. Longest Common Subsequence

Given two strings `text1`

and `text2`

, return *the length of their longest common subsequence. *If there is no

**common subsequence**, return

`0`

.A **subsequence** of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

- For example,
`"ace"`

is a subsequence of`"abcde"`

.

A **common subsequence** of two strings is a subsequence that is common to both strings.

**Example 1:**

Input:text1 = "abcde", text2 = "ace"Output:3Explanation:The longest common subsequence is "ace" and its length is 3.

**Example 2:**

Input:text1 = "abc", text2 = "abc"Output:3Explanation:The longest common subsequence is "abc" and its length is 3.

**Example 3:**

Input:text1 = "abc", text2 = "def"Output:0Explanation:There is no such common subsequence, so the result is 0.

**Constraints:**

`1 <= text1.length, text2.length <= 1000`

`text1`

and`text2`

consist of only lowercase English characters.

## Rust Solution

```
struct Solution;
impl Solution {
fn longest_common_subsequence(text1: String, text2: String) -> i32 {
let s1: Vec<char> = text1.chars().collect();
let s2: Vec<char> = text2.chars().collect();
let n1 = s1.len();
let n2 = s2.len();
let mut dp: Vec<Vec<usize>> = vec![vec![0; n2 + 1]; n1 + 1];
for i in 0..n1 {
for j in 0..n2 {
if s1[i] == s2[j] {
dp[i + 1][j + 1] = dp[i][j] + 1;
} else {
dp[i + 1][j + 1] = dp[i][j + 1].max(dp[i + 1][j]);
}
}
}
dp[n1][n2] as i32
}
}
#[test]
fn test() {
let text1 = "abcde".to_string();
let text2 = "ace".to_string();
let res = 3;
assert_eq!(Solution::longest_common_subsequence(text1, text2), res);
let text1 = "abc".to_string();
let text2 = "abc".to_string();
let res = 3;
assert_eq!(Solution::longest_common_subsequence(text1, text2), res);
let text1 = "abc".to_string();
let text2 = "def".to_string();
let res = 0;
assert_eq!(Solution::longest_common_subsequence(text1, text2), res);
}
```

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