word-break-ii（字典元素拼接单词）

题目描述

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"].

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        vector<string>res;
        if(dict.find(s)!=dict.end()) res.push_back(s);
        int size=s.size();
        for(int i=1;i<size;++i)
        {
            string word=s.substr(i);
            if(dict.find(word)==dict.end()) continue;
            vector<string>temp=wordBreak(s.substr(0,i),dict);
            Combine(temp,word);
            res.insert(res.begin(),temp.begin(),temp.end());
        }
        return res;
    }
private:
    void Combine(vector<string>&tt,const string app)
    {
        for(auto &elem:tt)
            elem+=' '+app;
        return ;
    }
};

拓展：字符串函数

#include <iostream>
#include<string>
using namespace std;

int main()
{
    string str="This is a just example.";
    cout<<str.substr(5)<<endl;
    size_t pos=str.find("just");
    cout<<str.substr(pos)<<endl;
    cout<<str.substr(0,3)<<endl;
    return 0;
}

运行结果如下：

is a just example.
just example.
Thi

Process returned 0 (0x0)   execution time : 0.068 s
Press any key to continue.

说明str.substr(n)表示从字符串str里截取从str[n]-str[str.size()-1]的元素的子串

而str.substr(m,n)表示从str[m]开始截取n个字符的子串

类似题目

题目描述

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s ="leetcode",
dict =["leet", "code"].

Return true because"leetcode"can be segmented as"leet code".

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        if(dict.find(s)!=dict.end()) return true;
        int size=s.size();
        for(int i=size-1;i>0;--i)
        {
            string word=s.substr(i);
            if(dict.find(word)==dict.end()) 
                continue;
            if(wordBreak(s.substr(0,i),dict)) return true;
        }
        return false;
    }
};