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  • 916. Word Subsets

    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<String> 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<String> 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:

    https://leetcode.com/problems/word-subsets/discuss/175854/C%2B%2BJavaPython-Straight-Forward

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10798047.html
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