LeetCode:Word Break II

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"]

.
 

class Solution {
public:

vector<string> myString;

vector<string> wordBreak(string s, unordered_set<string> &dict) {
<span style="white-space:pre">	</span>vector<string> result;
result.clear();
if(s.empty()||dict.empty())
return result;
bool* dp = new bool[s.size() + 1];
map<int,vector<int>> index;
memset(dp,false,s.size() + 1);
dp[0] = true;

for(int i=1;i<s.size()+1;i++)
{
<span style="white-space:pre">	</span>for(int j=0;j<i;j++)
{
if(dp[j])
{
string str = s.substr(j,i-j);
dp[i] = (dict.find(str) != dict.end())?true:false | dp[i];
if(dict.find(str) != dict.end())
{
map<int,vector<int>>::iterator it = index.find(i);
if(it!=index.end())
it->second.push_back(j);
else
{
vector<int> temp;
temp.push_back(j);
index.insert(make_pair(i,temp));
}
continue;
}
}
}
}

if(!dp[s.size()])
return result;
else
report(s,index,result,s.size());

return result;
}

void report(string &s,map<int,vector<int>>& index, vector<string>& result,int n)
{
if(n == 0)
{
string str;
for(int i = myString.size()-1;i>=0;i--)
{
str+=myString[i];
if(i!=0)
str.push_back(' ');
}
result.push_back(str);

}
else
{
map<int,vector<int>>::iterator it = index.find(n);
if(it!=index.end())
{
for(int i = 0; i<it->second.size();i++)
{
myString.push_back(s.substr(it->second[i],n - it->second[i]));
report(s,index,result,it->second[i]);
myString.pop_back();
}

}
}
}
};

继word break，首先用word break的方法判断是否可以匹配，然后利用保存的路径返回结果。

路径的保存用到了map，map的索引为字符串的字符对应的下标，索引对应的值为一个数组，该数组保存着以该索引为末尾字符的在dict中的匹配串的开始字符对应的下标。

例：

s = "abcd"

dict = ["a","b","abc","cd"]

map最终为

[

1->{0}

2->{1}

3->{0}

4->{2}

]

这样，当输出结果时，以尾字符开始递归查询，具体如下：

map.find(4)->second[0]=2   <----------------->  s.substr(2,4-2) = "cd"

map.find(2)->second[0]=1   <----------------->  s.substr(1,2-1) = "b"

map.find(1)->second[0]=0   <----------------->  s.substr(0,1-0) = "a"

将“a”、“b”、“cd”组合即可

已AC 52ms