472.Concatenated Words
===

###### tags: `Hard`,`Array`,`String`,`DP`,`DFS`,`Trie`

[472. Concatenated Words](https://leetcode.com/problems/concatenated-words/)

### 題目描述
Given an array of strings `words` (**without duplicates**), return *all the **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.

### 範例

**Example 1:**

Input: words = ["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".

**Example 2:**

Input: words = ["cat","dog","catdog"]
Output: ["catdog"]

**Constraints**:

* 1 <= `words.length` <= 10^4^
* 1 <= `words[i].length` <= 30
* `words[i]` consists of only lowercase English letters.
* All the strings of `words` are **unique**.
* 1 <= `sum(words[i].length)` <= 10^5^

### 解答
#### C++
##### 思路
1. 用DP檢查每個單字是否能用除了自己之外的單字來組成

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        unordered_set<string> s(words.begin(), words.end());
        vector<string> ans;
        for (auto word : words) {
            int n = word.size();
            vector<bool> dp(n + 1, false);
            dp[0] = true;
            for (int i = 0; i < n; i++) {
                if (!dp[i]) continue;
                for (int j = i + 1; j <= n; j++) {
                    if (j - i < n && s.count(word.substr(i, j - i))) {
                        dp[j] = true;
                    }
                }
                if (dp[n]) {
                    ans.push_back(word);
                    break;
                }
            }
        }
        return ans;
    }
};

> [name=Yen-Chi Chen][time=Mon, Jan 30, 2023]

### Reference

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