LeetCode - Word Break

Posted 2021-01-11 IncredibleThings

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.
Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".
Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.
Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

BFS需要一个队列来实现。首先根据在dict中查找s的前缀，如果有，加入队列中，作为遍历的“根”节点。比如上述的第二个例子，先入队的有"car"和"ca"两项；

当队列不为空时，队头top出列，令一个临时字符串temp是从s与top匹配后的字符开始到结束；如果此时temp是空，说明已经匹配完了，直接返回true，如果不为空，则进一步在dict中查找temp的前缀，如果有，加入队列中。

当队列为空且没有返回true时，说明匹配不成功，返回false。

按照这种做法，例子2首先入队"car"和"ca"，第一个出队的是"car"，temp是"s"，在dict中查找不到字符串"s"，就没有新的字符串入队；下一个出队的是"ca"，那么temp是"rs"，在dict中找到"rs"入队，下一步"rs"出队后，temp是空，返回true。

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        if(s == null || s.length() == 0){
            return false;
        }
        Queue<String> queue = new LinkedList<>();
        int[] visitedLength = new int[s.length()+1];
        for(int i = 0; i < wordDict.size(); i++){
            if(s.indexOf(wordDict.get(i)) == 0){
                queue.offer(wordDict.get(i));
                visitedLength[wordDict.get(i).length()] = -1;
            }
        }
        while(!queue.isEmpty()){
            String temp = queue.poll();
            if(temp.length() == s.length()){
                return true;
            }
            //temp always start from beginning of s
            String rest = s.substring(temp.length());
            for(int i = 0; i < wordDict.size(); i++){
                if(rest.indexOf(wordDict.get(i)) == 0 && visitedLength[(temp+wordDict.get(i)).length()] != -1){
                    queue.offer(temp+wordDict.get(i));
                    visitedLength[(temp+wordDict.get(i)).length()] = -1;
                }
            }
            
            
        }
        return false;
    }
}