472. Concatenated Words

最新推荐文章于 2020-12-14 21:35:40 发布

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本文介绍了一个程序，该程序接收一个无重复单词的列表，并返回所有由至少两个较短单词组成的组合词。通过深度优先搜索(DFS)算法实现，有效利用剪枝优化性能。

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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:

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

DFS求解，中间加入剪枝，程序如下所示：

class Solution {
    Set<String> set = new HashSet<>();
    List<String> list = new ArrayList<>();
    Map<String, Integer> map = new HashMap<>();
    private int cnt = 0;
    public List<String> findAllConcatenatedWordsInADict(String[] words) {    
        for (String val : words){
            set.add(val);
        }
        for (String val : words){
            cnt = 0;
            dfs(val, val.length(), 0, 0);
            if (cnt >= 2){
                list.add(val);
                map.put(val, cnt);
            }
        }
        return list;
    }
    public void dfs(String s, int sLen, int begin, int cur){
        if (begin > sLen){
            return ;
        }
        if (begin == sLen ){
            cnt = Math.max(cur, cnt);
            return;
        }
        if (cnt >= 2){
            return;
        }
        for (int i = begin; i < sLen; ++ i){
            if (i+1 <= sLen){
                String st = s.substring(begin, i+1);
                if (set.contains(st)){
                    String str = s.substring(i+1, sLen);
                    if (map.containsKey(str)){
                        cnt = cur + map.get(str);
                        return;
                    }
                    dfs(s, sLen, i + 1, cur+1);
                }
            }
        }
    }
}