LeetCode-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"

.

方法1 kmp+动归 复杂度O(n*L) n为字符串长度，L为字典大小

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 maxLen=0;
for(unordered_set<string>::iterator it=dict.begin();it!=dict.end();it++){
if( (*it).length()>maxLen)maxLen=(*it).length();
}
string tmp;
vector<bool>acc;
acc.resize(s.length()+1,false);
acc[0]=true;
for(int i=0;i<s.length();i++){
if(acc[i]){
int top=min(maxLen,(int)s.length()-i);
for(int j=1;j<=top;j++){
tmp=s.substr(i,j);
if(dict.find(tmp)!=dict.end()){
acc[i+j]=true;
}
}
}
}
return acc[s.length()];
}
};

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