zoukankan      html  css  js  c++  java
  • 472. Concatenated Words

    Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.

    A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

    Example:

    Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
    
    Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
    
    Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; 
    "dogcatsdog" can be concatenated by "dog", "cats" and "dog";
    "ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

    Note:

    1. The number of elements of the given array will not exceed 10,000
    2. The length sum of elements in the given array will not exceed 600,000.
    3. All the input string will only include lower case letters.
    4. The returned elements order does not matter.

    Approach #1: C++.

    class Solution {
    public:
        vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
            unordered_set<string> dict;
            vector<string> ans;
            sort(words.begin(), words.end(), [](string &a, string &b) { return a.length() < b.length(); });
            
            for (string word : words) {
                if (judge(word, dict)) 
                    ans.push_back(word);
                dict.insert(word);
            }
            
            return ans;
        }
        
    private:
        bool judge(string &word, unordered_set<string> &dict) {
            if (word.empty()) {
                return false;
            }
            
            int len = word.length();
            vector<bool> dp(len+1, false);
            dp[0] = true;
            
            for (int i = 1; i <= len; ++i) {
                for (int j = 0; j < i; ++j) {
                    if (dp[j] && dict.find(word.substr(j, i-j)) != dict.end()) {
                        dp[i] = true;
                        break;
                    }
                }
            }
            
            return dp[len];
        }
    };
    

      

    Analysis:

    dp[j] : the substring of word.substr(0, j) can be constituted with the word in dict.

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    寄存器英文全称
    汇编指令介绍
    汇编指令的基本知识
    第一篇
    Windows下让Git记住用户名密码(https)
    javascript 汉字拼音排序
    KO.js学习笔记(一)
    学javascript突发奇想,只用浏览器就能转换进制
    谨此纪念我的技术成长之路
    委托与事件
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10127387.html
Copyright © 2011-2022 走看看