LeetCode.720 Longest Word in Dictionary

题目：

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].分析1（推荐）：

class Solution {    public String longestWord(String[] words) {        //给定字符串数组，找出包含最长的字符串（包含其他单词的字母），如果存在相同长度，取字母排序小的        //思路：对数组排序，再利用Set对字母存储，小的单词一定包含在后面大的单词里面。后面只需要取前缀相同的                Set<String> set=new HashSet<String>();        for(int i=0;i<words.length;i++){            set.add(words[i]);        }        int length=0;        String word="";        for(int i=0;i<words.length;i++){            if(words[i].length()>length||(words[i].length()==length&&words[i].compareTo(word)<0)){                //如果存在相同长度的字符串，取字母排序较小的                int len=words[i].length();                while(len>0&&set.contains(words[i].substring(0,len))){                    //求相同的部分                    len--;                }                if(len==0){                    //说明该单词的所有字符串均为公共字母,将其标记为匹配串                    length=words[i].length();                    word=words[i];                }            }        }        return word;    }}

分析2:

class Solution {    public String longestWord(String[] words) {        //给定字符串数组，找出包含最长的字符串（包含其他单词的字母），如果存在相同长度，取字母排序小的        //思路：对数组排序，再利用Set对字母存储，小的单词一定包含在后面大的单词里面。后面只需要取前缀相同的                //对字母排序后，第一个单词一定是共有的，后面只需在此基础上添加         Arrays.sort(words);                HashSet<String> set=new HashSet<String>();        String res="";        for(String s:words){            //如果单词只有一个字母，那一定是共有的            if(s.length()==1||set.contains(s.substring(0,s.length()-1))){                res=s.length()>res.length()?s:res;                set.add(s);            }        }        return res;     }}