LeetCode - Word Break

標籤：div   input   size   leetcode   tco   style   bec   遍曆   節點   

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: trueExplanation: Return true because "leetcode" can be segmented as "leet code".Example 2:Input: s = "applepenapple", wordDict = ["apple", "pen"]Output: trueExplanation: 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需要一個隊列來實現。首先根據在dict中尋找s的首碼，如果有，排入佇列中，作為遍曆的“根”節點。比如上述的第二個例子，先入隊的有"car"和"ca"兩項；

當隊列不為空白時，隊頭top出列，令一個臨時字串temp是從s與top匹配後的字元開始到結束；如果此時temp是空，說明已經匹配完了，直接返回true，如果不為空白，則進一步在dict中尋找temp的首碼，如果有，排入佇列中。

當隊列為空白且沒有返回true時，說明匹配不成功，返回false。

按照這種做法，例子2首先入隊"car"和"ca"，第一個出隊的是"car"，temp是"s"，在dict中尋找不到字串"s"，就沒有新的字串入隊；下一個出隊的是"ca"，那麼temp是"rs"，在dict中找到"rs"入隊，下一步"rs"出隊後，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;    }}

 

LeetCode - Word Break