word-break
2019-05-10 本文已影响0人
cherryleechen
时间限制:1秒 空间限制:32768K
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".
我的代码
class Solution {
public:
bool wordBreak(string s, unordered_set<string> &dict) {
int siz=s.size();
if(dict.find(s)!=dict.end())
return true;
for(int i=1;i<siz;i++){
string word=s.substr(i);
if(dict.find(word)==dict.end())
continue;
if(wordBreak(s.substr(0,i),dict))
return true;
}
return false;
}
};
运行时间:5ms
占用内存:704k
class Solution {
public:
bool wordBreak(string s, unordered_set<string> &dict) {
int siz=s.size();
vector<bool> dp(siz+1,false);
dp[0]=true;//记录s.substr(0,i)是否可以分词
for(int i=1;i<=siz;i++)
for(int j=0;j<i;j++)
if(dp[j]&&(dict.find(s.substr(j,i-j))!=dict.end())){
dp[i]=true;
break;
}
return dp[siz];
}
};
运行时间:6ms
占用内存:608k