zoukankan      html  css  js  c++  java
  • LeetCode 1048. Longest String Chain

    原题链接在这里:https://leetcode.com/problems/longest-string-chain/

    题目:

    Given a list of words, each word consists of English lowercase letters.

    Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2.  For example, "abc" is a predecessor of "abac".

    word chain is a sequence of words [word_1, word_2, ..., word_k] with k >= 1, where word_1 is a predecessor of word_2word_2 is a predecessor of word_3, and so on.

    Return the longest possible length of a word chain with words chosen from the given list of words.

    Example 1:

    Input: ["a","b","ba","bca","bda","bdca"]
    Output: 4
    Explanation: one of the longest word chain is "a","ba","bda","bdca".

    Note:

    1. 1 <= words.length <= 1000
    2. 1 <= words[i].length <= 16
    3. words[i] only consists of English lowercase letters.

    题解:

    Sort the words based on word length.

    For each word, find all its possible predecessors by deleting one char from the word, if it exist, update longest chain length up to this word. 

    And put it in the mapping.

    At the same time, update the global longest res.

    Time Complexity: O(nlogn + n*len). n is the words count. len is the average length of word.

    Space: O(n).

    AC Java:

     1 class Solution {
     2     public int longestStrChain(String[] words) {
     3         HashMap<String, Integer> hm = new HashMap<>();
     4         Arrays.sort(words, (a, b) -> a.length() - b.length());
     5         
     6         int res = 0;
     7         for(String word : words){
     8             int best = 0;
     9             for(int i = 0; i<word.length(); i++){
    10                 String prev = word.substring(0, i) + word.substring(i+1);
    11                 best = Math.max(best, hm.getOrDefault(prev, 0)+1);
    12             }
    13             
    14             hm.put(word, best);
    15             res = Math.max(res, best);
    16         }
    17         
    18         return res;
    19     }
    20 }
  • 相关阅读:
    为什么会需要消息队列(MQ)?
    RBAC用户角色权限设计方案
    转:jquery 父、子页面之间页面元素的获取,方法的调用
    LeetCode Wiggle Subsequence
    LeetCode Longest Arithmetic Sequence
    LeetCode Continuous Subarray Sum
    LeetCode Maximum Length of Repeated Subarray
    LeetCode Is Subsequence
    LeetCode Integer Break
    LeetCode Largest Sum of Averages
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11750286.html
Copyright © 2011-2022 走看看