标签：leetcode
Word Break && Word Break II
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".
dp[j]&&substr(j,i-j) 此时的j 从 0 到 i-1
拿上面的leetcode来说，当 i=4，此时表示leet单词是否在字典中存在，因为 dp[0]&&substr(0,4-0)
此时substr(0,4-0)=leet 在字典中存在，所以 dp[4]=true,依次往后求，直到求的dp[size].
代码：
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int size=s.size();
vector<bool> dp(s.length() + 1, false);
dp[0]=true;
for(int i=1;i<=size;i++)
{
string temp;
for(int j=0;j<i;j++)
{
if(dp[j]&&wordDict.find(s.substr(j,i-j))!=wordDict.end())
{
dp[i]=true;
break;
}
}
}
return dp[size];
}
};
Word Break II
代码如下：
class Solution {
public:
vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
vector<string> result;
int len=s.size();
string temp;
for(int i=0;i<len;i++)
{
temp=s.substr(0,i+1);
if(wordDict.find(temp)!=wordDict.end())
{
findNext(s,i+1,temp,wordDict,result);
}
temp="";
}
return result;
}
void findNext(string s,int start,string temp,unordered_set<string>& wordDict,vector<string>&result)
{
if(start>=s.length())
{
result.push_back(temp);
return;
}
string now;
for(int i=start;i<s.length();i++)
{
now=s.substr(start,i-start+1);
if(wordDict.find(now)!=wordDict.end())
{
findNext(s,i+1,temp+‘ ‘+now,wordDict,result);
}
}
}
};
下面有一中优化做法，参考博客:Word berak Ⅱ
1 使用二维表vector<vector<int> >tbl记录，记录从i点，能否跳到下一个break位置。如果不能，那么tbl[i]就为空。如果可以，就记录可以跳到哪些位置。
2 利用二维表优化递归回溯法。
优化点：
如果当前位置是start，但是tbl[start]为空，那么就是说，这个break位置不能break整个s串的，直接返 回上一层，不用搜索到下一层了。
class Solution {
public:
//2014-2-19 update
vector<string> wordBreak(string s, unordered_set<string> &dict)
{
vector<string> rs;
string tmp;
vector<vector<int> > tbl = genTable(s, dict);
word(rs, tmp, s, tbl, dict);
return rs;
}
void word(vector<string> &rs, string &tmp, string &s, vector<vector<int> > &tbl,
unordered_set<string> &dict, int start=0)
{
if (start == s.length())
{
rs.push_back(tmp);
return;
}
for (int i = 0; i < tbl[start].size(); i++)
{
string t = s.substr(start, tbl[start][i]-start+1);
if (!tmp.empty()) tmp.push_back(‘ ‘);
tmp.append(t);
word(rs, tmp, s, tbl, dict, tbl[start][i]+1);
while (!tmp.empty() && tmp.back() != ‘ ‘) tmp.pop_back();//tmp.empty()
if (!tmp.empty()) tmp.pop_back();
}
}
vector<vector<int> > genTable(string &s, unordered_set<string> &dict)
{
int n = s.length();
vector<vector<int> > tbl(n);
for (int i = n - 1; i >= 0; i--)
{
if(dict.count(s.substr(i))) tbl[i].push_back(n-1);
}
for (int i = n - 2; i >= 0; i--)
{
if (!tbl[i+1].empty())//if we can break i->n
{
for (int j = i, d = 1; j >= 0 ; j--, d++)
{
if (dict.count(s.substr(j, d))) tbl[j].push_back(i);
}
}
}
return tbl;
}
};
class Solution {
public:
//2014-2-19 update
vector<string> wordBreak(string s, unordered_set<string> &dict)
{
vector<string> rs;
string tmp;
vector<vector<int> > tbl = genTable(s, dict);
word(rs, tmp, s, tbl, dict);
return rs;
}
void word(vector<string> &rs, string tmp, string &s, vector<vector<int> > &tbl,
unordered_set<string> &dict, int start=0)
{
if (start == s.length())
{
rs.push_back(tmp);
return;
}
for (int i = 0; i < tbl[start].size(); i++)
{
string t = s.substr(start, tbl[start][i]-start+1);
if (!tmp.empty())
{
word(rs, tmp+‘ ‘+t, s, tbl, dict, tbl[start][i]+1);
}
else
{
word(rs, t, s, tbl, dict, tbl[start][i]+1);
}
}
}
vector<vector<int> > genTable(string &s, unordered_set<string> &dict)
{
int n = s.length();
vector<vector<int> > tbl(n);
for (int i = n - 1; i >= 0; i--)
{
if(dict.count(s.substr(i))) tbl[i].push_back(n-1);
}
for (int i = n - 2; i >= 0; i--)
{
if (!tbl[i+1].empty())//if we can break i->n
{
for (int j = i, d = 1; j >= 0 ; j--, d++)
{
if (dict.count(s.substr(j, d))) tbl[j].push_back(i);
}
}
}
return tbl;
}
};
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标签：leetcode
原文地址：http://blog.csdn.net/yujin753/article/details/48010677
