LeetCode: Word Break II [140]

【题目】

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"].

【题意】

给定一个字符串s和词典dict, 返回全部切分情况。使得切分后每一个单词都是dict中的单词

【思路】

依次确定以每一个位置i结尾的单词的前驱单词集合（仅仅要记住前驱单词的结束位置）

        然后从后往前恢复切分路径就可以。

DP问题

【代码】

class Solution {
public:

    void recoverPath(vector<string>&result, vector<string>&words, string&s, map<int, vector<int> >preorders, int end){
        vector<int>preorder=preorders[end];
        for(int i=0; i<preorder.size(); i++){
            string word=s.substr(preorder[i]+1, end-preorder[i]);
            if(preorder[i]==-1){
                string cutstr=word;
                int size=words.size();
                for(int j=size-1; j>=0; j--){
                    cutstr+=" "+words[j];
                }
                result.push_back(cutstr);
            }
            else{
                words.push_back(word);
                recoverPath(result, words, s, preorders, preorder[i]);
                words.pop_back();
            }
        }
    }

    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        vector<string> result;
        if(s.length()==0)return result;
        map<int, vector<int> >preorders;    //记录各个可分位置的前驱集合
        vector<int> pos(1,-1);         //以确定可分的位置

        for(int i=0; i<s.length(); i++){
            vector<int>preorder;
            for(int k=0; k<pos.size(); k++){
                if(dict.find(s.substr(pos[k]+1, i-pos[k]))!=dict.end()){
                    preorder.push_back(pos[k]);
                }
            }
            if(preorder.size()>0){
                preorders[i]=preorder;
                pos.push_back(i);
            }
        }

        //恢复全部可能的切分路径
        if(preorders.find(s.length()-1)==preorders.end())return result;
        vector<string>words;
        recoverPath(result, words, s, preorders, s.length()-1);
        return result;
    }
};