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916. Word Subsets
阅读量:432 次
发布时间:2019-03-06

本文共 2056 字,大约阅读时间需要 6 分钟。

We are given two arrays A and B of words.  Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity.  For example, "wrr" is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is universal if for every b in Bb is a subset of a

Return a list of all universal words in A.  You can return the words in any order.

 

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]Output: ["facebook","leetcode"]

 

Note:

  1. 1 <= A.length, B.length <= 10000
  2. 1 <= A[i].length, B[i].length <= 10
  3. A[i] and B[i] consist only of lowercase letters.
  4. All words in A[i] are unique: there isn't i != j with A[i] == A[j].

 

Approach #1: String. [Java]

class Solution {    public List
wordSubsets(String[] A, String[] B) { int[] init = new int[26], temp; for (String b : B) { temp = counter(b); for (int i = 0; i < 26; ++i) { init[i] = Math.max(init[i], temp[i]); } } List
ret = new ArrayList<>(); for (String a : A) { temp = counter(a); int i = 0; for (i = 0; i < 26; ++i) { if (temp[i] < init[i]) break; } if (i == 26) ret.add(a); } return ret; } public int[] counter(String b) { int[] temp = new int[26]; for (int i = 0; i < b.length(); ++i) { temp[b.charAt(i)-'a']++; } return temp; }}

  

Reference:

 

转载地址:http://gftuz.baihongyu.com/

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