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524.Longest-Word-in-Dictionary-through-Deleting

524. Longest Word in Dictionary through Deleting
题目地址
​https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/​
题目描述
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:
All the strings in the input will only contain lower-case letters.
The size of the dictionary won't exceed 1,000.
The length of all the strings in the input won't exceed 1,000.
代码
Approach #1 Brute Force
Time: O(2^n) && Space: O(2^n)
class Solution {
  public String findLongestWord(String s, List<String> d) {
        HashSet<String> set = new HashSet<>(d);
    List<String> l = new ArrayList<>();
    generate(s, "", 0, l);
    String max_str = "";
    for (String str: l) {
      if (set.contains(str)) {
        if (str.length() > max_str.length() 
            || (str.length() == max_str.length())
           && str.compareTo(max_str) < 0) {
          max_str = str;
        }
      }
    }
    return max_str;
  }
​
  public void generate(String s, String str, int i, List<String> l) {
    if (i == s.length()) {
      l.add(str);
    } else {
      generate(s, str + s.charAt(i), i + 1, l);
      generate(s, str, i + 1, l);
    }
  }
}
Approach #2 Iterative Brute Force
Time: O(2^n) && Space: O(2^n)
class Solution {
  public String findLongestWord(String s, List<String> d) {
    HashSet<String> set = new HashSet<>(d);
    List<String> l = new ArrayList<>();
    for (int i = 0; i < (1 << s.length()); i++) {
      String t = "";
      for (int j = 0; j < s.length(); j++) {
        if (((i >> j) & 1) != 0) {
          t += s.charAt(j);
        }
      }
      l.add(t);
    }
    String max_str = "";
    for (String str: l) {
      if (set.contains(str)) {
        if (str.length() > max_str.length() 
            || (str.length() == max_str.length())
           && str.compareTo(max_str) < 0) {
          max_str = str;
        }
      }
    }
    return max_str;
  }
}
Approach #3 Sorting
class Solution {
  public String findLongestWord(String s, List<String> d) {
    Collections.sort(d, (a, b) -> a.length() != b.length() ? 
                     b.length() - a.length() : a.compareTo(b));
    for (String dictWord: d) {
      int i = 0;
      for (char c: s.toCharArray()) {
        if (i < dictWord.length()
           && c == dictWord.charAt(i)) {
          i++;
        }
      }
      if (i == dictWord.length())
        return dictWord;
    }
​
    return "";
  }
}
Approach #4 Without Sorting
class Solution {
  public String findLongestWord(String s, List<String> d) {
    String longest = "";
    for (String dictWord: d) {
      int i = 0;
      for (char c: s.toCharArray()) {
        if (i < dictWord.length() && c == dictWord.charAt(i)) {
          i++;
        }
      }
      if (i == dictWord.length() && dictWord.length() >= longest.length()) {
        if (dictWord.length() > longest.length() || dictWord.compareTo(longest) < 0) {
          longest = dictWord;
        }
      }
    }
    return longest;
  }
}
Approach #5 Sorting and Checking Subsequence
Time: O(n * x long n + n * x)
class Solution {
  public boolean isSubsequence(String x, String y) {
    int j = 0;
    for (int i = 0; i < y.length() && j < x.length(); i++) {
      if (x.charAt(j) == y.charAt(i)) {
        j++;
      }
    }
    return j == x.length();
  }
​
  public String findLongestWord(String s, List<String> d) {
    Collections.sort(d, new Comparator<String> () {
      public int compare(String s1, String s2) {
        return s2.length() != s1.length() ?
          s2.length() - s1.length() : s1.compareTo(s2);
      }
    });
​
    for (String str: d) {
      if (isSubsequence(str, s)) {
        return str;
      }
    }
​
    return "";
  }
}

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Last modified 3yr ago