LWC 57：720. Longest Word in Dictionary

LWC 57：720. Longest Word in Dictionary

传送门：720. Longest Word in Dictionary

Problem:

Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.

If there is no answer, return the empty string.

Example 1:

Input:
words = [“w”,”wo”,”wor”,”worl”, “world”]
Output: “world”
Explanation:
The word “world” can be built one character at a time by “w”, “wo”, “wor”, and “worl”.

Example 2:

Input:
words = [“a”, “banana”, “app”, “appl”, “ap”, “apply”, “apple”]
Output: “apple”
Explanation:
Both “apply” and “apple” can be built from other words in the dictionary. However, “apple” is lexicographically smaller than “apply”.

Note:

• All the strings in the input will only contain lowercase letters.
• The length of words will be in the range [1, 1000].
• The length of words[i] will be in the range [1, 30].

思路：
对于候选的答案如”apple”，从字典集中找出”a”,”ap”,”app”,”appl”，如果均能找到则进行最大长度的比较。为了长度最大，且字典序最小，当遇到相同字符串长度时，还需要再比较一次字符大小。

代码如下：

    public String longestWord(String[] words) {        int n = words.length;        Set<String> set = new HashSet<>();        set.add("");        for (int i = 0; i < n; ++i){            set.add(words[i]);        }        String ans = "";        int maxLen = 0;        for (int i = 0; i < n; ++i){            String word = words[i];            if (valid(word, set)){                int len = word.length();                if (len >= maxLen){                    if (len == maxLen){                        if (ans.compareTo(word) > 0){                            ans = word;                        }                    }                    else{                        ans = word;                    }                    maxLen = len;                }            }                    }        return ans;    }    boolean valid(String word, Set<String> set){        int n = word.length();        for (int i = 0; i < n; ++i){            if (!set.contains(word.substring(0, i))) return false;        }        return true;    }