Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

这题是Word Break的follow up。现在不是判断能否被break,而是给出所有break的情况。如果用单纯的search＋backtrack会TLE，这种做法如下：

class Solution(object):
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: Set[str]
        :rtype: List[str]
        """
        if not s or not wordDict:
            return []
        maxlen = 0
        for w in wordDict:
            maxlen = max(maxlen, len(w))
        hash = set(wordDict)
        res = []
        self.search(s, hash, res, [], 0, maxlen)
        res = self.gen_result(res, s)
        return res
        
    def search(self, s, hash, res, cur, index, maxlen):
        if index == len(s):
            res.append(cur + [index])
            return 
        cur.append(index)
        for i in xrange(1,maxlen+1):
            if s[index:index+i] in hash:
                self.search(s, hash, res, cur, index+i, maxlen)
        cur.pop()
    
    def gen_result(self, res, s):
        tmp = []
        for i in xrange(len(res)):
            arr = []
            for j in xrange(0,len(res[i])-1):
                arr.append(s[res[i][j] : res[i][j+1]])
            tmp.append(' '.join(arr))
        return tmp

如何改进呢～为何搜索复杂度高，主要在于重复分割，如果后面需要分割的子段前面曾经分割过，则再次处理时就比较浪费时间。一个以空间换时间的做法是将之前处理过的子串的所有可能都保存，是一种memory search的方法。以空间换取了时间。代码如下：

class Solution(object):
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: Set[str]
        :rtype: List[str]
        """
        #pure search and backtracking is TLE, using the memory search method,
        #ie, a dict object to store the result for the current tackling substr
        cache = {}
        result = self.helper(s, wordDict, cache)
        return result
        
    def helper(self, s, wordDict, cache):
        if s in cache:
            return cache[s]
        result = []
        if s in wordDict:
            result.append(s)
        for i in xrange(1, len(s)):
            word = s[i:]
            if word in wordDict:
                rem = s[:i]
                prev = self.combine(word, self.helper(rem, wordDict, cache))
                result += prev
        cache[s] = result + []
        return result
    
    def combine(self, word, prev):
        #word is a string, and prev is a list of string，
        tmp = []
        for i in xrange(len(prev)):   
            tmp.append(prev[i] + ' ' + word) ＃注意这段一定要这样写，防止直接对prev的修改造成内存的直接修改，影响到cache.
        return tmp

 

转载于:https://www.cnblogs.com/sherylwang/p/5774470.html