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"
.
基本思路:
動态規劃。
數組 dp[i],表示第i個字元以前是否可以分隔成單詞。 i 從0開始。
已知dp[0..i]時,求dp[i+1],則需要償試,s[k..i], 0<=k <=i, 進行償試。
在leetcode上實際執行時間為12ms。
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
vector<bool> dp(s.size()+1);
dp[0] = true;
for (int i=1; i<=s.size(); i++) {
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[s.size()];
}
};