Word Break
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"
才发现我一直把题意理解错了，原来是源串分出来的词只要都在字典里就好了，比如leetleet就可以分成两个leet都在词典里，原来不是所有字典里的词都要出现。
那就容易些了，另dp[i] 表示源串的前i个字符可以满足分割，那么 dp[ j ] 满足分割的条件是存在k 使得 dp [k] && substr[k,j]在字典里。
这是O(n2)的算法，注意下标，因为很恶心。
class Solution {
public:
bool wordBreak(string s, unordered_set<string> &dict) {
// Note: The Solution object is instantiated only once and is reused by each test case.
int n = (int)s.size();
vector<bool> dp(n + 1, false);
dp[0]=true;
for(int i=1;i<=n;i++)
{
if(dp[i-1])
{
int idx=i-1;
for(int j=idx;j<n;j++)
{
string cur=s.substr(idx,j-idx+1);
if(dict.count(cur)>0)
dp[j+1]=true;
}
}
}
return dp[n];
}
};
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"].
跟第一题一样，就是要返回所有可能的切分， 做切分反应是用回溯，但是不剪枝肯定要超时。
这里用了一个math[i][j] 来表示 i--j 这一段是否可以切分，然后在dfs的时候利用看最后剩余的子串能否切分来剪枝。
class Solution {
public:
vector<string> wordBreak(string s, unordered_set<string> &dict)
{
int n=s.length();
vector<vector<bool> > match(n+1,vector<bool>(n+1,false));
for(int i=0;i<=n;i++)
match[0][i]=true;
for(int len=1;len<=n;len++)
{
for(int start=0;start+len<=n;start++)
{
string tmp=s.substr(start,len);
if(dict.count(tmp)>0)
match[len][start]=true;
else
{
for(int left=1;left<len;left++)
{
match[len][start]=match[left][start]&&match[len-left][start+left];
if(match[len][start])
break;
}
}
}
}
if(match[n][0]==false)
return vector<string>();
vector<string> ans;
vector<string> had;
dfs(s,0,match,had,ans,dict);
return ans;
}
void dfs(string& s,int k,vector<vector<bool> >& match,vector<string>& had,vector<string>& ans,unordered_set<string> &dict)
{
int n=s.length();
if(k>=n)
{
if(!had.empty())
{
string ret;
for(int i=0;i<had.size();i++)
{
ret.append(had[i]);
if(i!=had.size()-1)
ret.push_back(' ');
}
ans.push_back(ret);
return;
}
}
for(int len=1;k+len<=n;len++)
{
string tmp=s.substr(k,len);
if(dict.count(tmp)>0 && match[n-k-len][k+len])
{
had.push_back(tmp);
dfs(s,k+len,match,had,ans,dict);
had.pop_back();
}
}
}
};