Leetcode---Word Break
2015-01-02 12:01
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Given a string s[/i] and a dictionary of words dict[/i], determine if s[/i] can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s[/i] = "leetcode",
dict[/i] = ["leet", "code"].
Return true because "leetcode" can be segmented as "leet code".
Have you been asked this question in an interview?
此题相当简单，动态规划
bool wordBreak(string s, unordered_set<string> &dict) {
int len=s.length();
vector<bool> dp(len+1,false);
dp[0]=true;
for(int i=0;i<=len;i++){
for(int j=0;j<i;j++){
if(dict.find(s.substr(j,i-j))!=dict.end() && dp[j]==true){
dp[i]=true;
}
}
}
return dp[len];
}
For example, given
s[/i] = "leetcode",
dict[/i] = ["leet", "code"].
Return true because "leetcode" can be segmented as "leet code".
Have you been asked this question in an interview?
此题相当简单，动态规划
bool wordBreak(string s, unordered_set<string> &dict) {
int len=s.length();
vector<bool> dp(len+1,false);
dp[0]=true;
for(int i=0;i<=len;i++){
for(int j=0;j<i;j++){
if(dict.find(s.substr(j,i-j))!=dict.end() && dp[j]==true){
dp[i]=true;
}
}
}
return dp[len];
}
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