Word Break Ⅱ

问题描述

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

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"].

算法分析

1、从字符串开始处分析，如果第一个单词是cat，则该组剩余的单词是在sanddog 里提取，即sand和dog ；

2、如果第一个单词是cats ，则该组剩余的单词在在anddog 里提取，即and和dog ；

3、很明显这是一个递归问题，找出一个单词后递归查找剩余的单词。

4、使用一个unsorted_map<string, vector<string>> 数据结构存储每一次递归查找一个字符串的结果。查找"cat" 的结果为["cat"] ；查找"cats" 的结果为["cats"] ；查找"catsand" 的结果为["cat sand", "cats and"] ；查找"catsanddog" 的结果为["cat sand dog", "cats and dog"] 。

C++实现

class Solution {private：    unordered_map<string, vector<string>> m;    vector<string> combine(string word, vector<string> prev) {        for(int i = 0; i < prev.size();++i)            prev[i] += " " + word;        return prev;    }    bool isaWord(string s, vector<string>& wordDict) {        vector<string>::iterator it = find(wordDict.begin(), wordDict.end(), s);        return (it != wordDict.end());    }public:    vector<string> wordBreak(string s, vector<string>& wordDict) {        if (m.count(s))             return m[s]; //take from memory        vector<string> result;        if (isaWord(s, wordDict))  //a whole string is a word            result.push_back(s);        for(int i = 1;i < s.size(); ++i) {            string word = s.substr(i);            if(isaWord(word, wordDict)) {                string rem = s.substr(0, i);                vector<string> prev = combine(word, wordBreak(rem, wordDict));                result.insert(result.end(), prev.begin(), prev.end());            }        }        m[s] = result; //memorize        return result;    }};