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.

 

这道题给了一个由单词组成的数组,某些单词是可能由其他的单词组成的,让我们找出所有这样的单词。这道题跟之前那道Word Break十分类似,我们可以对每一个单词都调用之前那题的方法,我们首先把所有单词都放到一个unordered_set中,这样可以快速找到某个单词是否在数组中存在。对于当前要判断的单词,我们先将其从set中删去,然后调用之前的Word Break的解法,具体讲解可以参见之前的帖子。如果是可以拆分,那么我们就存入结果res中,参见代码如下:

 

解法一:

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        if (words.size() <= 2) return {};
        vector<string> res;
        unordered_set<string> dict(words.begin(), words.end());
        for (string word : words) {
            dict.erase(word);
            int len = word.size();
            if (len == 0) continue;
            vector<bool> v(len + 1, false);
            v[0] = true;
            for (int i = 0; i < len + 1; ++i) {
                for (int j = 0; j < i; ++j) {
                    if (v[j] && dict.count(word.substr(j, i - j))) {
                        v[i] = true;
                        break;
                    }
                }
            }
            if (v.back()) res.push_back(word);
            dict.insert(word);
        }
        return res;
    }
};

 

下面这种方法跟上面的方法很类似,不同的是判断每个单词的时候不用将其移除set,而是在判断的过程中加了判断,使其不会判断单词本身是否在集合set中存在,而且由于对单词中子字符串的遍历顺序不同,加了一些优化在里面,使得其运算速度更快一些,参见代码如下:

 

解法二:

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

 

下面这种方法是递归的写法,其中递归函数中的cnt表示有其他单词组成的个数,至少得由其他两个单词组成才符合题意,参见代码如下:

 

解法三:

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        vector<string> res;
        unordered_set<string> dict(words.begin(), words.end());
        for (string word : words) {
            if (word.empty()) continue;
            if (helper(word, dict, 0, 0)) {
                res.push_back(word);
            }
        }
        return res;
    }
    bool helper(string& word, unordered_set<string>& dict, int pos, int cnt) {
        if (pos >= word.size() && cnt >= 2) return true;
        for (int i = 1; i <= (int)word.size() - pos; ++i) {
            string t = word.substr(pos, i);
            if (dict.count(t) && helper(word, dict, pos + i, cnt + 1)) {
                return true;
            }
        }
        return false;
    }
};

 

类似题目:

Word Break

 

参考资料:

https://discuss.leetcode.com/topic/72393/c-772-ms-dp-solution

https://discuss.leetcode.com/topic/72433/c-600ms-20-lines-of-code-dfs-solution-is-there-any-way-to-optimize

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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