LeetCode每日一题(524. Longest Word in Dictionary through Deleting)

Given a string s and a string array dictionary, return the longest string in the dictionary that can be formed by deleting some of the given string characters. If there is more than one possible result, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

 

Example 1:

Input: s = "abpcplea", dictionary = ["ale","apple","monkey","plea"]
Output: "apple"

Example 2:

Input: s = "abpcplea", dictionary = ["a","b","c"]
Output: "a"

 

Constraints:

• 1 <= s.length <= 1000
• 1 <= dictionary.length <= 1000
• 1 <= dictionary[i].length <= 1000
• s and dictionary[i] consist of lowercase English letters.

 

代码(Rust):

impl Solution {
    pub fn find_longest_word(s: String, dictionary: Vec<String>) -> String {
        // 依据s生成一个二位数组， 记录每一个字母出现的index
        let mut c_to_i: Vec<Vec<i32>> = vec![vec![]; 26];
        for (i, c) in s.chars().enumerate() {
            c_to_i[c as usize - 97].push(i as i32);
        }
        let mut ans = String::new();
        'outer: for word in dictionary {
            let mut prev_pos = -1;
            for c in word.chars() {
                // 如果字母没有在s中出现则直接检查下一个word
                if c_to_i[c as usize - 97].is_empty() {
                    continue 'outer;
                } else {
                    // 找第一个大于prev_pos的位置
                    if let Some(&pos) = c_to_i[c as usize - 97].iter().find(|&v| v > &prev_pos) {
                        // 如果存在则把前一个字符index更新为当前的index
                        prev_pos = pos;
                    } else {
                        // 如果不存在则直接下一个单词
                        continue 'outer;
                    }
                }
            }
            if word.len() > ans.len() || (word.len() == ans.len() && word < ans) {
                ans = word;
            }
        }
        ans
    }
}