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:
动态规划。
数组 dp[i],表示第i个字符以前是否可以分隔成单词。 i 从0开始。
已知dp[0..i]时,求dp[i+1],则需要偿试,s[k..i], 0<=k <=i, 进行偿试。
在leetcode上实际执行时间为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()];
}
};