字符串断开问题（word break II） （LeetCode--递归篇） 难

给定一个字符串s和一个单词词典，在s中添加空格来构造一个句子，每个单词都是一个有效的字典单词。返回所有可能的句子。例如，给定s ="catsanddog"， dict =["cat"， "cats"， "and"， "sand"， "dog"]。解决办法是“cats and dog”、“cat sand dog”。

1、暴力递归

解题思路：

如果从0到i的字符串是一个单词，则继续使用rest子字符串进行计算

递归过程中，若当前已到最后一个字母，则压入结果容器中

超时！！！

cpp文件

#include "stdafx.h";
#include "restoreIPAddresses.h";
using namespace std;

vector<string> Solution::wordBreak(string s, unordered_set<string> &dict) {
vector<string> res;
string list = "";
subbreak(s, list, dict, 0, res);
return res;
};
void Solution::subbreak(string s, string list, unordered_set<string>& dict, int len, vector<string>& res){
if (len == s.length()) {
res.push_back(list);
}
for (int j = 1; j <= s.length() - len; j++){
string s1 = s.substr(len, j);
unordered_set<string>::const_iterator got = dict.find(s1);
if (got == dict.end())
continue;
if (list == "")
list = s1;
else{
list = list + " ";
list += s1;
}
subbreak(s, list, dict, len + j, res);
list = "";
}

}

h文件

#include "stdafx.h";
using namespace std;
class Solution {
public:
static vector<string> wordBreak(string s, unordered_set<string> &dict);
static void subbreak(string s, string list, unordered_set<string>& dict, int len, vector<string>& res);
};

main文件

#include "stdafx.h"
#include "restoreIPAddresses.h";
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
string s = "catsanddog";
unordered_set<string> dict = { "cat", "cats", "and", "sand", "dog" };
Solution::wordBreak(s, dict);


return 0;

}

stdafx.h文件

#pragma once


#include "targetver.h"


#include <stdio.h>
#include <tchar.h>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include <hash_set>
#include <iostream>
#include <algorithm>
#include <sstream>

#include <unordered_set>

2、动态规划

参考：https://github.com/summer00072/leetcode/blob/master/algorithms/cpp/wordBreak/wordBreak.II.cpp

//---------------------
	// Dynamic Programming
	//---------------------
	//
	// Define substring[i, j] is the sub string from i to j.
	//
	// (substring[i,j] == word) : result[i] = substring[i,j] + {result[j]}
	//
	// So, it's better to evaluate it backword.
	//
	// For example:
	//
	// s = "catsanddog",
	// dict = ["cat", "cats", "and", "sand", "dog"].
	//
	// 0 c "cat" -- word[0,2] + {result[3]} ==> "cat sand dog"
	// "cats" -- word[0,3] + {result[4]} ==> "cats and dog"
	// 1 a ""
	// 2 t ""
	// 3 s "sand" -- word[3,6] + {result[7]} ==> "sand dog"
	// 4 a "and" -- word[4,6] + {result[7]} ==> "and dog"
	// 5 n ""
	// 6 d ""
	// 7 d "dog"
	// 8 o ""
	// 9 g ""
	vector<string> wordBreak_dp(string s, set<string> &dict) {
	vector< vector<string> > result(s.size());
	for(int i=s.size()-1; i>=0; i--) {
	vector<string> v;
	result[i] = v;
	for(int j=i+1; j<=s.size(); j++) {
	string word = s.substr(i, j-i);
	if (dict.find(word) != dict.end()){
	if (j==s.size()){
	result[i].push_back(word);
	}else{
	for(int k=0; k<result[j].size(); k++){
	result[i].push_back(word + " " + result[j][k]);
	}
	}
	}
	}
	}
	return result[0];
	}

牛客网未完全通过

原因是顺序的问题