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
Rust Solution
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"];
assert_eq!(Solution::word_break(s, word_dict), true);
let s = "applepenapple".to_string();
let word_dict = vec_string!["apple", "pen"];
assert_eq!(Solution::word_break(s, word_dict), true);
let s = "catsandog".to_string();
let word_dict = vec_string!["cats", "dog", "sand", "and", "cat"];
assert_eq!(Solution::word_break(s, word_dict), false);
}
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