524. Longest Word in Dictionary through Deleting

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

Example 1:

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

Output:
"apple"

Example 2:

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

Output:
"a"

Note:

1. All the strings in the input will only contain lower-case letters.
2. The size of the dictionary won't exceed 1,000.
3. The length of all the strings in the input won't exceed 1,000.

 public class Solution {
     public String findLongestWord(String s, List<String> d) {
         Comparator<String> comp = new Comparator<String>(){
           public int compare(String s1,String s2){
               if(s1.length()!=s2.length()){
                   return s2.length()-s1.length();
               }else{
                   return s1.compareTo(s2);
               }
           }
         };
         Collections.sort(d,comp);
         for(String str:d){
             int i=0;
             for(char c:s.toCharArray()){
                 if(i<str.length()&&str.charAt(i)==c) i++;
             }
             if(i==str.length()) return str;
         }
         return "";
     }
 }
 //the run time complexity could be nlongn, the space complexity could be O(1);

 public class Solution {
     public String findLongestWord(String s, List<String> d) {
         String longest = "";
         for(String str:d){
             int i = 0;
             for(char c:s.toCharArray()){
                 if(i<str.length()&&str.charAt(i)==c) i++;
             }
             if(i==str.length()&&str.length()>=longest.length()){
                 if(str.length()>longest.length()||str.compareTo(longest)<0){
                     longest = str;
                 }
             }
         }
         return longest;
     }
 }
 //the run time could be O(nkaveragep),space complexity could be O(1);