139. Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

```Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because `"leetcode"` can be segmented as `"leet code"`.
```

Example 2:

```Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because `"`applepenapple`"` can be segmented as `"`apple pen apple`"`.
Note that you are allowed to reuse a dictionary word.
```

Example 3:

```Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false
```

139. Word Break
``````struct Solution;
use std::collections::HashSet;
use std::iter::FromIterator;

impl Solution {
fn word_break(s: String, word_dict: Vec<String>) -> bool {
let n = s.len();
let mut a = vec![false; n + 1];
let hs: HashSet<String> = HashSet::from_iter(word_dict);
a[0] = true;
for i in 1..=n {
for j in 0..i {
if a[j] && hs.contains(&s[j..i]) {
a[i] = true;
break;
}
}
}
a[n]
}
}

#[test]
fn test() {
let s = "leetcode".to_string();
let word_dict = vec_string!["leet", "code"];
let res = true;
assert_eq!(Solution::word_break(s, word_dict), res);
let s = "applepenapple".to_string();
let word_dict = vec_string!["apple", "pen"];
let res = true;
assert_eq!(Solution::word_break(s, word_dict), res);
let s = "catsandog".to_string();
let word_dict = vec_string!["cats", "dog", "sand", "and", "cat"];
let res = false;
assert_eq!(Solution::word_break(s, word_dict), res);
}
``````